Diff [PDF]

Zitiervorschau

Department of Mathematical Sciences

MA1002 Calculus Differential Calculus Dr John Pulham

ii

September 13, 1999, Version 1.3 Copyright  1999 by Ian Craw and the University of Aberdeen All rights reserved. Additional copies may be obtained from: Department of Mathematical Sciences University of Aberdeen Aberdeen AB9 2TY DSN: mth199-101462-2

Contents 1 The 1.1 1.2 1.3 1.4

Differential Calculus Introduction . . . . . . . . . . . . . . . . . . . . . . . . . The slope of a graph and the tangent . . . . . . . . . . Increasing and Decreasing . . . . . . . . . . . . . . . . . Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Rate of Change . . . . . . . . . . . . . . . . . . . 1.5 Some Basic Derivatives . . . . . . . . . . . . . . . . . . . 1.5.1 f (x) = xn where n is a positive integer. . . . . 1 . . . . . . . . . . . . . . . . . . . . . 1.5.2 f (x) = x 1.5.3 The Absolute Value f (x) = |x| . . . . . . . . . . 1.5.4 Derivatives of sin x and cos x . . . . . . . . . . 1.6 Rules of Differentiation . . . . . . . . . . . . . . . . . . . 1.6.1 Rules . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Examples of the Use of the Rules . . . . . . . . . . . . . 1.7.1 Polynomials . . . . . . . . . . . . . . . . . . . . . n = 1, 2, 3, 4 . . . 1.7.2 The function f (x) = x,n 1.7.3 Long Products . . . . . . . . . . . . . . . . . . . 1.7.4 Trigonometric Functions . . . . . . . . . . . . . . 1.7.5 Rational Functions . . . . . . . . . . . . . . . . . √ 1.7.6 The square root f (x) = x . . . . . . . . . . . 1.8 Derivative of sin x where x is in radians . . . . . . . . 1.9 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . 1.10 Proofs of some of the Rules . . . . . . . . . . . . . . . . 1.11 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . 1.11.1 Geometrical Interpretation of Second Derivative . 1.11.2 Kinematical Interpretation . . . . . . . . . . . . . 1.12 Limits & Continuity . . . . . . . . . . . . . . . . . . . . 1.12.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . 1.12.1 Continuity . . . . . . . . . . . . . . . . . . . . . . 1.12.2 * Rolle’s Theorem and the Mean Value Theorem . iii

. . . . . . .

1 1 1 3 3 4 5 5

. . . . . . . . . .

6

. . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

6 7 8 8 9 9 10 10 10 11 12 14 17 22 23 23 24 26 26 27 28

iv

CONTENTS

2 Further Differentiation 2.1 The Inverse Trigonometric Functions . . . . . . . . 2.1.1 Inverse Functions in general . . . . . . . . . 2.1.1 The Inverse Trigonometric Functions . . . . 2.1.2 Differentiating the Inverse Trig Functions . 2.1.3 Polar Coordinates . . . . . . . . . . . . . . . 2.2 Implicit Functions and their Differentiation . . . . 2.2.1 Differentiating Implicitly Defined Functions 2.3 Parametric Representation . . . . . . . . . . . . . 2.3.1 Differentiation in Parametric Form . . . . . 3 The Exponential and Logarithm Functions 3.1 Exponential Population Growth . . . . . . . 3.2 The Exponential Function . . . . . . . . . . . 3.2.1 Definition . . . . . . . . . . . . . . . . 3.2.2 Properties of the Exponential Function 3.2.3 Summary . . . . . . . . . . . . . . . . 3.2.4 Differentiation Examples . . . . . . . . 3.3 The Natural Logarithm . . . . . . . . . . . . 3.3.1 Definition of the Logarithm . . . . . . 3.3.2 Derivative of ln x . . . . . . . . . . . . 3.4 Hyperbolic Functions . . . . . . . . . . . . . 3.4.1 More Differentiation Examples . . . . . 3.5 Powers . . . . . . . . . . . . . . . . . . . . . . 3.6 * Irrational Numbers . . . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

31 31 31 32 34 35 38 39 42 43

. . . . . . . . . . . . .

47 47 48 49 51 52 53 54 55 56 56 57 59 60

A Some Useful Tables

63

B Solutions to Exercises

65

List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

The tangent at a point . . . . . . . . . . . . . . . . . . . . The derivative of x2 from first principles. . . . . . . . . . . The graph of y = |x|. . . . . . . . . . . . . . . . . . . . . . The derivative of sin x from first principles. . . . . . . . . . Second derivative shape . . . . . . . . . . . . . . . . . . . Second derivative shape . . . . . . . . . . . . . . . . . . . The graph of a step function. . . . . . . . . . . . . . . . . A discontinuity. . . . . . . . . . . . . . . . . . . . . . . . . Rolle: somewhere between a and b, the graph must be flat. Mean Value Theorem . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

2 3 6 15 24 24 28 28 29 30

2.1 2.2 2.3 2.4 2.5

Graph of sin x. . . . . . . . . . . . Graph of arcsin(x). . . . . . . . . . Graph of arccos(x). . . . . . . . . . Polar co-ordinates (r, θ) of a point. The lemniscate of Example 2.6. . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

33 34 34 36 41

3.1 3.2 3.3

Population growth behaviour. . . . . . . . . . . . . . . . . . . . . . . . . . Graph of ex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graph of ln(x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

48 52 55

v

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

vi

LIST OF FIGURES

How to use these Notes The notes are divided into • theory and explanation • worked examples • exercises Most of the theory will also be presented in lectures. The exceptions are the sections marked with a ∗. These are ‘extras’ and are not an examinable part of the course. They are not necessarily more difficult than the official bits—it’s just that they are not in the syllabus. Read them if you are interested. The more you read of everything the more you are likely to understand. Please study the worked examples carefully because that is usually the best way to grasp the theory. There is also a lot to be said for treating some of the worked examples as exercises in the first instance, trying to solve them yourself and only then reading through my solution. It is usually rather silly to try reading mechanically through a worked example when you have not yet got into your head what the example is really about. The exercises come in two varieties—unstarred and starred. The starred exercises are either a bit more difficult than the others or else are less directly relevant to the course. The more exercises you try the better, but do not get worried if you have not attempted many of the starred questions. The standard of the examination questions is based on the unstarred questions. There are some solutions and some answers at the back of the book. Use these sensibly. Make a serious attempt at a problem before looking up the answer or you will just be wasting your time.

vii

viii

LIST OF FIGURES

Chapter 1 The Differential Calculus 1.1

Introduction

The Calculus, comprising the Differential Calculus and the Integral Calculus, is one of the great achievements of Mathematics. Some aspects of it date back to the ancient Greeks but we normally regard it as the brain-child of Isaac Newton (1642–1727), with help from Gottfried Leibniz (1646–1716) and a push start from people like Pierre Fermat and Barrow (get used to the fact that Mathematics is resolutely international!). The basic problem that led up to the development of the Differential Calculus was that of constructing the tangent line to an arbitrary curve at a given point. But, for Newton, the much more serious problem was that of describing motion and making rigorous the concepts of velocity and acceleration. The Calculus solved this problem for him and with it Newton was able to construct his System of the World that has dominated our thought ever since. Over the last few centuries the Calculus has grown enormously to be a dominant influence on the development of Mathematics itself and an indispensable tool in the application of Mathematics to Science, Engineering, Biology, Economics etc. The Calculus is in the remarkable position of being intellectually profound, aesthetically beautiful and very practical all at the same time. There are very few other human achievements of which this can be said with the same force. This course does not pretend to give you more than a brief introduction to the Calculus, but it does contain most of the basic ideas that propel the subject.

1.2

The slope of a graph and the tangent

One of the problems that gave rise to the calculus was that of finding the equation of the tangent to a curve. The fundamental idea of the calculus is to obtain this information—the slope of the graph—by a limiting process. 1

2

CHAPTER 1. THE DIFFERENTIAL CALCULUS

The argument goes as follows. Suppose we want to find the slope of the graph y = f (x) at the point P(a, f (a)) on the graph. We take another point Q(a + h, f (a + h)) close to P on the graph (either side) and work out the slope of the chord PQ. This comes out as f (a + h) − f (a) f (a + h) − f (a) = (1.1) (a + h) − a h

y=f(x)

Q

tangent P f(a+h) f(a) x a

a+h

We call this the Newton Quotient. Figure 1.1: The tangent at a point Now for the critical step. We imagine Q moving closer and closer to P (h → 0). As it does so it seems clear that the slope of the line PQ will get closer and closer to the slope of the tangent line at P. So, we claim that the slope of the tangent line at P is given by the limiting value of (1.1) as h shrinks down to zero. This value, the slope of the tangent at P, is called the derivative of f (x) at x = a df f (a + h) − f (a) (a) = lim h→0 dx h

(1.2)

Everything in the differential calculus comes from this. The clever, and difficult, thing about this definition is its use of a limiting process. We shall have a little more to say about limits later on but for the moment note that taking the limit as h → 0 is not the same thing as putting h = 0 in the formula. If you put h = 0 in the Newton Quotient you just get 0/0—which does not mean anything (Newton’s critics had a lot to say about that!). Note that we frequently use the alternative notation f 0 (a) for the derivative. Example 1.1. The derivative of f (x) = x2 Let me do an example at once so as to clarify matters. Let’s find the slope of the graph y = x2 at the point (2, 4) on the graph. P is the point (2, 4) and Q is the point (2 + h, (2 + h)2 ) (NOT (2 + h, 4 + h)). The slope of the chord PQ is given by the Newton Quotient (2 + h)2 − 22 f (2 + h) − f (2) = h h We want to find the limiting value of this slope as h → 0. Once more note that we cannot just put h = 0 in the formula. Now 4h + h2 (2 + h)2 − 4 = =4+h h h It is now clear that as we let h shrink to zero the value of the Newton Quotient tends to 4 as its ultimate limit. So the slope of the graph at (2, 4) is 4.

1.3. INCREASING AND DECREASING

3

y

y=x 2 Q P

x 2

2+h

Figure 1.2: The derivative of x2 from first principles. More generally, if we take the point P(a, a2 ) on the graph the Newton Quotient becomes 2ah + h2 (a + h)2 − a2 = = 2a + h h h and as h → 0 this slope tends to the value 2a. So we have shown that the value of the derivative of f (x) = x2 at x = a is 2a, or f 0 (x) = 2x

1.3

Increasing and Decreasing

The derivative f 0 (x) of f (x) gives the slope of the graph y = f (x) at the point (x, f (x))—i.e. the slope of the tangent line to the graph at this point. You already know that a positive slope of a line means that the line is sloping upwards and that a negative slope of a line means that it is sloping downwards. This shows us the following important facts: if f 0 (x) > 0 then the value of f (x) is increasing if f 0 (x) < 0 then the value of f (x) is decreasing There is one other important point to be brought out here: The derivative of a constant function is zero and if f 0 (x) is always zero then f (x) is constant.

1.4

Velocity

The other main impetus to the invention of the calculus was the study of dynamics. In particular, the clarification of the concept of speed (or velocity).

4

CHAPTER 1. THE DIFFERENTIAL CALCULUS

If you ask a car driver how fast he is driving he will look at his speedometer and say something like ‘50 mph’. What does this mean? It obviously means something, or he wouldn’t have said it. It does not mean that he will travel 50 miles in the next hour. His speed is, presumably, changing all the time. What he is trying to say is that, at this particular moment, he is doing 50 mph. This is the idea that we have to clarify. The problem can be approached like this. Suppose, for simplicity, that he is moving along the x-axis and is in position x(t) at time t (x is a function of t). At a slightly later time t + h he is in position x(t + h). So, in time h he has covered distance x(t + h) − x(t) (assuming that he has not backed up in the meantime). So we can reasonably say that his average speed between times t and t + h is x(t + h) − x(t) distance = time h which is the Newton Quotient again! But we don’t want his average speed over the time from t to t + h, we want his speed at time t. The next step is predictable. We define his speed at time t to be the limiting value of this average speed as h → 0. x(t + h) − x(t) h→0 h

speed at time t = lim

The right-hand side of this is just a standard Newton Quotient. This shows us that if a particle is moving along the x-axis and its x coordinate at time t is x(t) the the velocity of the particle at time t is given by the derivative of x with respect to t. dx velocity = dt Notes: 1. I use the technical term velocity rather than speed. Speed is the magnitude of velocity, without the sign. Velocities can be negative, speeds cannot. 2. For historical reasons it is common to use the notation x˙ for a derivative with respect to time. 3. Try explaining all this to the driver.

1.4.1

Rate of Change

In general, if P (t) is a quantity that depends on time t then its rate of change is defined to be its derivative with respect to time. You have to be a bit careful about signs here. If the volume of an object at time t is V (t) and I say that the volume is increasing at a rate α

1.5. SOME BASIC DERIVATIVES

5

then I am saying that V˙ (t) = α. If, on the other hand, I say that the volume is decreasing at a rate α then I am saying that V˙ (t) = −α. Be careful to distinguish between rates of increase and decrease.

1.5

Some Basic Derivatives

Let us now have a look at the derivatives of some other functions.

1.5.1

f (x) = xn where n is a positive integer.

The Newton Quotient is (x + h)n − xn f (x + h) − f (x) = . h h How do we work out the limiting value of this as h → 0? As usual, there is nothing to be learned from just putting h = 0 in the RHS because, as usual, we just get 0/0. The complicated bit on the RHS is the term (x + h)n . Here are a few cases expanded out: (x + h)2 = x2 + 2hx + h2 , (x + h)3 = x3 + 3hx2 + 3h2 x + h3 , (x + h)4 = x4 + 4hx3 + 6h2 x2 + 4h3 x + h4 . In general we have n (x + h) factors multiplied together. If we think of what we are going to get when we multiply out these brackets we see that (1) if we take an x from each bracket that gives us a xn term, (2) if we take a h from one bracket and x’s from all the others that gives us a xn−1 h term. We could have chosen any of the n brackets to give the h so, all told, we get nxn−1 h. (3) any other term that we are going to get when multiplying out the product is going to contain at least a h2 . That is all that we need to know about the expansion. We have argued that (x + h)n = xn + nhxn−1 + terms involving at least h or h2 or ... or hn . So (x + h)n − xn = nhxn−1 + terms involving at least h or h2 or ... or hn . So the Newton Quotient can be written as (x + h)n − xn = nxn−1 + terms involving at least h. h As h → 0 all these extra terms tend to zero. So: (x + h)n − xn lim = nxn−1 . h→0 h So we have the result that d n x = nxn−1 dx

for

n = 1, 2, 3, 4, . . .

6

1.5.2

CHAPTER 1. THE DIFFERENTIAL CALCULUS

f (x) =

1 x

Proceed as usual. The Newton Quotient is 1 f (x + h) − f (x) = h h



 1 1 − . x+h x

Simplifying the RHS by algebra we get f (x + h) − f (x) −1 = . h x(x + h) Now we let h tend to zero. It is clear that, unless x = 0, the RHS is tending to the value −1/x2 , because (x + h) is tending to x. So we have the result that, if x 6= 0,   1 d 1 = − 2. dx x x

1.5.3

The Absolute Value f (x) = |x|

|x| is what is known as the absolute value of x. It is defined by ( x if x ≥ 0, |x| = −x if x < 0. You can describe this as: if x is negative then throw away the negative sign. So, for example, |3| = 3, |π| = π, | − 3| = 3, | − π| = π. You can easily check for yourself, by keeping track of positive and negative signs, that y |xy| = |x||y|

|xn | = |x|n

A less obvious property is that known as the Triangle Inequality |x + y| ≤ |x| + |y|

y = |x|

x

Figure 1.3: The graph of y = |x|. The graph of y = |x| looks like Fig 1.3. For negative x it is the line y = −x and for positive x it is the line y = x. Note that it turns through a right-angle at the origin. The question “What is the slope of y = |x| at the origin?” does not seem to have a sensible answer. So we ought to find ourselves in difficulties if we try to differentiate f (x) = |x| at x = 0. Let’s see what happens. The Newton Quotient is |x + h| − |x| f (x + h) − f (x) = h h

1.5. SOME BASIC DERIVATIVES

7

We are interested in x = 0, so the Newton Quotient becomes |h| |h| − 0 = h h What is the limiting value of this as h → 0? I said nothing in my original definition of the derivative to suggest that h had to be positive. We have to allow for the possibility of h tending to zero through negative values, or even a combination of positive and negative values. Now look at what happens in this case. If h is tending to zero through positive values we have, at all times, h |h| = =1 h h So the limiting value as h → 0 is 1. If h is tending to zero through negative values we have, at all times, −h |h| = = −1 h h So the limiting value as h → 0 is −1. So, in this case the Newton Quotient does NOT have a unique limiting value as h → 0. In such a case we say that the function f (x) does not have a derivative at this point, or that f (x) is a non-differentiable function (at this point). In conclusion, it is fairly easy to see that the function f (x) = |x| is differentiable for x 6= 0 with derivative −1 if x < 0 and derivative 1 if x > 0 and is not differentiable at x = 0. (Non-differentiability normally shows up as a kink or break in the graph, where we cannot make sense of the tangent.)

1.5.4

Derivatives of sin x and cos x

The derivatives of these two basic trig functions are d sin x = cos x dx

d cos x = − sin x dx

It is very important to remember two things about these. First, remember the negative sign in the derivative of the cosine. Second, remember that the formulas only work if the angle x is measured in radians! We will come back to prove these results after doing the next section.

8

CHAPTER 1. THE DIFFERENTIAL CALCULUS

Questions 1

(Hints and solutions start on page 65.)

Q 1.1. Go through the proof that the derivative of x3 is 3x2 , starting from the definition of the derivative. Q 1.2. Find the derivative of 1/(1 + x) from the definition. Q 1.3. The derivative of sin x is cos x. So the value of the derivative of sin x at x = 0 should be 1. Use your calculator to evaluate the Newton Quotient (sin(0 + h) − sin(0))/h for h = 0.1, 0.05, 0.01 and so on and compare with the limiting result. Remember to switch your calculator to radians! *Q 1.4. Draw a picture to illustrate the following statement. P is the point (x, f (x)) on the graph of y = f (x). R is the point (x + h, f (x + h)) and Q is the point (x − h, f (x − h)). The Newton Quotient is the slope of the chord PR. The limiting value of this slope as h → 0 gives the slope of the graph at P. On the other hand the slope of the chord QR seems to give a much better approximation to the slope of the graph at P and should give the value of the slope in the limit as h → 0. 1 (f (x + h) − f (x − h)) and try some examples Show that the slope of QR is given by 2h to compare this approximation to the derivative with that given by the Newton Quotient. Suppose I take the limiting value of this new expression as my new definition of f 0 (x). Give an example of a function that is differentiable according to this new definition, but not differentiable according to the proper definition. Can you give a proof that if a function is differentiable at x according to the proper definition then it will also be differentiable at x according to the new definition.

1.6

Rules of Differentiation

So far we have had to work out all our derivatives from the definition. If this were the only way to do differentiation the Calculus would never have got off the ground. Luckily, it turns out that once we have a few derivatives under our belt the rest come more-or-less for free. This is because there are certain mechanical rules for differentiating functions that are combinations of functions that we know how to differentiate.

1.6.1

Rules

I will state six of the rules here. Then I will do some examples. Finally I will state the seventh rule.

1.7. EXAMPLES OF THE USE OF THE RULES Suppose that f and g are differentiable functions and that λ λ0 (λf )0 (f + g)0 (f − g)0 (f g)0  0 f g

9 1

is a constant. Then

= 0 derivative of constant is zero = λf 0 can take out constants = f 0 + g 0 Sum Rule = f 0 − g 0 follows from [2] and [3] = f 0 g + f g 0 Product Rule =

f 0g − f g0 g2

Quotient Rule

[1] [2] [3] [4] [5] [6]

In the other notation we have, for example, df dg d (f g) = x.g + f. x. dx dx dx The rules are not independent of each other. You can easily check that [2] and [4] follow from the others. The rules tell us how to differentiate sums, differences, products and quotients of functions that we already know how to differentiate. The process can be used over and over again. For example, it follows immediately that (f + g + h)0 = (f + g)0 + h0 = f 0 + g 0 + h0 .

1.7 1.7.1

Examples of the Use of the Rules Polynomials

A polynomial of degree n is a function of the form p(x) = an xn + an−1 xn−1 + an−2 xn−2 + · · · + a1 x + a0 where the coefficients an , an−1 , . . . are constants. For example, 2x − 3, x4 − 2x + 5 or πx10 − 12 x5 + 2.343 are polynomials. Polynomials are easy to differentiate using the rule that we have produced for differentiating powers and rules [1], [2] and [3] above. Example 1.2. You will eventually differentiate polynomials in your head — they are that easy. But let me start by spelling the process out in great detail so as to show the way in which it uses the Rules. Consider the polynomial p(x) = x2 + 4x − 3. Using the Sum Rule (twice over) we get p0 (x) = 1

d d 2 d (x ) + (4x) + (−3). dx dx dx

For a discussion of greek letters, and a table of the alphabet, see the Appendix on page 64.

10

CHAPTER 1. THE DIFFERENTIAL CALCULUS

d We already know how to get derivatives of powers, so we know that dx (x2 ) = 2x. The d d rule on taking out constants tells us that dx (4x) = 4 dx (x) = 4. Finally, the derivative of a constant is zero. So we end up with

p0 (x) = 2x + 4. As I said, you will get to do that mentally in time. Example 1.3. Now for a few more, without all the intervening details. d 2 d d 2 d (x + 3x − 4) = (x ) + 3 (x) − (4) = 2x + 3, dx dx dx dx d 5 d 5 d 4 d 4 (x − 4x + 2x − 3) = x − 4 x + 2 x = 5x4 − 16x3 + 2, dx dx dx dx d (2 − 3x8 + x16 ) = −24x7 + 16x15 . dx

1.7.2

The function f (x) = x−n

n = 1, 2, 3, 4 . . .

Using the quotient rule: d n 0.xn − 1. dx x d 1 −nxn−1 d −n x = = = = −nx−n−1 dx dx xn (xn )2 x2n

So we now know that the rule d n x = nxn−1 dx works for all whole numbers n. In the same vein, if f (x) = 1/g(x) then f 0 (x) =

1.7.3

0.g(x) − 1.g 0(x) g 0 (x) = − . g(x)2 g(x)2

Long Products

This is an example of using the product rule repeatedly. Suppose that f , g and h are three differentiable functions and that p = f.g.h. Then, by the product rule: p0 = (f gh)0 = ((f g)h)0 = (f g)0h + (f g)h0 = (f 0 g + f g 0)h + f gh0 = f 0 gh + f g 0 h + f gh0.

1.7.4

Trigonometric Functions

We have already met sin and cos. The other standard trig functions are tan x =

sin x , cos x

cot x =

cos x , sin x

sec x =

1 , cos x

cosec x =

1 . sin x

1.7. EXAMPLES OF THE USE OF THE RULES

11

We can now use the rules (mainly the quotient rule) to differentiate these. d tan x = dx

d dx

d sin x. cos x − sin x dx cos x cos2 x + sin2 x 1 = = . 2 2 cos x cos x cos2 x

provided that x is measured in radians. So d tan x = sec2 x. dx Similarly

d 0. cos x − 1. dx cos x sin x tan x d sec x = = = . 2 2 dx cos x cos x cos x

So sin x d sec x = = sec x tan x. dx cos2 x Similarly cos x d cosec x = − 2 = − cosec x cot x. dx sin x and d cot x = − cosec2 x. dx I suggest that you get to know the derivatives of sine, cosine and tan. The others can be worked out as you need them.

1.7.5

Rational Functions

A Rational Function is a function of the form R(x) =

p(x) q(x)

where p(x) and q(x) are polynomials. Here are two examples of rational functions: R(x) =

x2 − 3x + 1 , x−3

R(x) =

3x − 5 . + x5 + 1

x10

Rational functions can be differentiated by using the quotient rule, since we know how to differentiate polynomials. R(x) =

p(x) q(x)

gives

R0 (x) =

p0 (x)q(x) − p(x)q 0 (x) . q(x)2

For example R(x) =

x2 + x + 1 x−3

R0 (x) =

(2x + 1)(x − 3) − 1.(x2 + x + 1) x2 − 6x − 4 = . (x − 3)2 (x − 3)2

12

CHAPTER 1. THE DIFFERENTIAL CALCULUS

Another example: R(x) =

1 − x2 x2 + x + 1

R0 (x) =

−2x(x2 + x + 1) − (1 − x2 )(2x + 1) , (x2 + x + 1)2

and I leave it to you to simplify the numerator.

The square root f (x) =

1.7.6

√

x

This can be done in a slightly indirect way2 as follows. Let g(x) = f (x).f (x) = x. Then, by the product rule, √ 1 = g 0 (x) = f 0 (x).f (x) + f (x).f 0 (x) = 2f (x)f 0 (x) = 2 xf 0 (x). Therefore

1 d√ x= √ , dx 2 x

or

1 1 d 1 x 2 = x 2 −1 . dx 2

We will see later that, in fact, d a x = axa−1 . dx for all values of a. Example 1.4. Differentiate f (x) = 4x sin x + 3 cos x. Here we start to put together almost everything we know. In order, we use the sum rule, followed by the product rule and end up using the fact that we know how to differentiate the trig functions. f 0 (x) = 4(1. sin x + x cos x) − 3 sin x = sin x + 4x cos x. Example 1.5. Differentiate f (x) = x sin x cos x. A function like this really needs the product rule twice over. f 0 (x) = 1.(sin x cos x) + x.

d (sin x cos x) = sin x cos x + x(cos x cos x − sin x sin x). dx

sin(x) + cos(x) . sin(x) − cos(x) This is, first of all, a quotient. So we start by using the Quotient Rule:

Example 1.6. Differentiate

0

f (x) = 2

d (sin x dx

f (x) =

d + cos x).(sin x − cos x) − (sin x + cos x) dx (sin x − cos x) . 2 (sin x − cos x)

A more direct way is given in Exercise 1.16

1.7. EXAMPLES OF THE USE OF THE RULES

13

We then do the remaining derivatives in the numerator and tidy up f 0 (x) =

(cos x − sin x)(sin x − cos x) − (sin x + cos x)(cos x + sin x) . (sin x − cos x)2

Now multiply out the numerator and show that it simplifies down to -2 and that the answer is therefore −2 f 0 (x) = (sin x − cos x)2

Questions 2

(Hints and solutions start on page 65.)

Q 1.5. Differentiate the following functions f (x) = x4 ,

1 g(x) = x5 + x − 3, 2

h(t) = 4t6 − 3t5 + 2t − 4,

k(p) = p4 + 3p−3 + 2p−1 .

Q 1.6. Now differentiate these p(x) = 2 sin x,

q(x) = tan x + sin x,

5

m(λ) = λ2 − 4 sin λ − 3 cos λ,

1

1

h(t) = 3t 4 − 6t 3 + t−2 ,

1

φ(x) = tan x + 3 cos x − x 2 .

Q 1.7. Differentiate the following functions using the product rule. f (x) = x sin x,

r(x) = cos2 x.

g(x) = cos x sin x,

p(x) = x(1 + 2 sin x),

q(x) = (sin x + cos x)(sin x − cos x).

u(x) = (tan x − 1)(tan x + 1),

5

v(x) = tan2 x,

w(t) = (1 + t 4 )(t2 − t3 ).

Q 1.8. Differentiate the following, starting with the quotient rule. p(x) =

1 , x

r(x) =

x2 + 1 u(x) = 2 , x −1

v(x) =

sin x , 1 + cos x

x2

x , +1

s(x) =

t2 + t + 1 e(t) = 2 , t −t+1

w(x) =

cos x , 1 − sin x

3x + 1 , 2x + 3

t(x) =

√ t−t m(t) = √ , t+t

z(x) =

1

f (x) = x 2 − x− 2 ,

r(θ) = 3 cos θ − 4 sin θ,

1 , sin x

2x + 1 . −x+1

x2

l(x) =

f (x) =

1 . 1 + x1

1 . cos x sin x

14

CHAPTER 1. THE DIFFERENTIAL CALCULUS

Q 1.9. Find the equation of the tangent line to the curve y = x3 at the point (1, 1). Where does the tangent meet the x-axis? Q 1.10. Find the equation of the tangent to the graph of y = sin(x) at the point where x = π. Find the points where this tangent meets the x and y-axes. Q 1.11. Let P be the point (a, a1 ) (a 6= 0) on the graph y = x1 where a > 0. Find the equation of the tangent line at P. Find where this line meets the x- and y-axes. What is the area of the triangle formed by the tangent and the two axes? What is surprising about the answer? Q 1.12. The Normal at a point P on a curve is the line through P that is perpendicular to the tangent line. What is the slope of the normal in terms of the slope of the tangent? Find the equation of the normal at the point (a, a2 ) on the curve y = x2 . Where does it meet the y-axis? Q 1.13. For what values of x is the function f (x) = x(1 − x)(2 − x) increasing and for what values is it decreasing? Draw a picture. Q 1.14. Consider the function f (x) = (x2 +1)/(x2 −1). For what values of x is this function defined? For what values of x is it increasing and for what values of x is it decreasing? Try to draw a picture in this case too. *Q 1.15. Sketch the graph of y = x2 . Find the equation of the tangent to this graph at the point P with coordinates (a, a2 ). Find the point Q on the graph at which the tangent is perpendicular to the tangent at P (a 6= 0). Show that the tangents at P and Q meet on the horizontal line y = − 14 . What is the corresponding result for a circle? *Q 1.16. Show that √ √ a−b √ . a− b= √ a+ b √ Use this to find the derivative of f (x) = x from the definition of the derivative using Newton Quotients.

1.8

Derivative of sin x where x is in radians

As promised (or threatened) earlier, we are now going to find out the derivative of sin(x) from first principles. Please remember that x is measured in radians. The Newton Quotient is sin(x + h) − sin x f (x + h) − f (x) = h h

1.8.

DERIVATIVE OF SIN X WHERE X IS IN RADIANS

15

What happens to this as h → 0? This turns out to be slightly difficult to answer. Once more we get nowhere by putting h = 0. We have to be cleverer than that. There is a trig formula that says     A−B A+B sin sin A − sin B = 2 cos 2 2 (you should prove this by expanding out the RHS) So sin(x + h) − sin x = 2 cos(x + h2 ) sin( h2 ) and hence Newton Quotient =

sin(x + h) − sin x h sin(h/2) = cos(x + ) · h 2 h/2

How does this help us? Let me start with the easy bit. As h → 0 the value of cos(x + h/2) as tends to cos x. It’s the other bit that gives problems. What happens to the ratio sin(h/2) h/2 h → 0? Let us try plugging in some values to see what we get

0.2

sin(h/2) h/2 0.99831

0.1

0.99958

0.01

0.9999958

h

This rather suggests that the ratio is tending to 1. That is indeed true and we will now prove it. Theorem 1.17. As θ → 0 then sin(θ)/θ → 1 (θ in radians). Proof. Consider Fig. 1.4, showing a sector of a circle of radius R and having angle θ (where θ is positive and less than a right-angle).

C

O

D

θ A B

Figure 1.4: The derivative of sin x from first principles. By trigonometry, OA = R cos θ, AC = R sin θ and BD = R tan θ. It is clear from the diagram that area 4 OAC < area of sector OBC < area 4 OBD

16

CHAPTER 1. THE DIFFERENTIAL CALCULUS

Or

1 1 1 .OA.AC < R2 θ < .OB.BD 2 2 2 Putting in the values that we have obtained (with OB = OC = R) we get 1 2 1 1 R sin θ cos θ < R2 θ < R2 tan θ 2 2 2 1 2 Now divide through by 2 R and get

sin θ cos θ We are thinking of θ as a small positive angle (!) so sin θ > 0. So we can divide through by sin θ and get θ 1 cos θ < < sin θ cos θ Now let θ → 0. The outer terms in this inequality are both tending to the value cos 0=1. The term in the middle is sandwiched between them at all times, so it must be tending to the limit 1 as well. So θ → 1 as θ → 0 sin θ and hence sin θ → 1 as θ → 0 θ Strictly speaking, we have only proved this for positive values of θ. But changing the sign of θ does not affect the value of sin(θ)/θ, so our argument works for negative values of θ as well. sin θ cos θ < θ < tan θ =

Now we go back to our original Newton Quotient h sin(h/2) sin(x + h) − sin(x) = lim cos(x + ) · = cos(x).1 = cos x lim h→0 h→0 h 2 h/2 So, so long as x is measured in radians d sin x = cos x dx In a similar way it can be shown that d cos x = − sin x dx Footnote: Look at the following argument. We know how to differentiate sin(x). We also know that sin2 x + cos2 x = 1. So, differentiating both sides of this equation wrt x, using the product rule on sin2 x = sin(x) sin(x) etc., we get 2 sin(x) cos(x) + 2 cos(x)

d (cos(x)) = 0. dx

Solving this for the derivative gives us d cos x = − sin x. dx The answer is right, but what do you think of the logic? Is it acceptable?

1.9. THE CHAIN RULE

1.9

17

The Chain Rule

This is the final rule of differentiation. It is a bit more complicated than the others. Let me start by saying something about functions. Up to now we have been treating functions as being, basically, formulas “y = x2 + 1”. A better way to think of a function like f (x) is as a machine, a piece of electronic wizardry, which takes in x as its input and gives out f (x) as its output. x → f → f (x) A formula, like f (x) = x2 + 1, is then just describing what the machine does. If x goes in x2 + 1 comes out. If α goes in α2 + 1 comes out. If 2 goes in 5 comes out. If x + 1 goes in (x + 1)2 + 1 comes out. All that we ask of a function is that it be ‘well-defined’, i.e. that the output is totally determined by the input. There is no necessity that the function should be given by a ‘formula’ in any straightforward sense. For later reference I should also point out that we sometimes need to put conditions on the type of input that is allowed — the domain of definition of the function. A function that is supposed to work out the smallest factor of a whole number is obviously not going to take too kindly to being given an input like 23.2342 because a number like this does not have ‘factors’. Similarly, a function that is supposed to be working out the square root of a number will not approve of a negative input. If we have two functions, f and g—two machines, we can think of forming a single machine by plugging the output of one into the input of the other (connecting them in series). g → f → f (g(x)) x → This new machine represents a function called the composition of f and g. Note that the order matters, the following is probably a different machine x

→

f

→ √

g

→

g(f (x))

So, for example, the function h(x) = 1 + x√2 can be thought of as the result of composing two functions f (x) = 1 + x2 and g(x) = x √ f → 1 + x2 → g → 1 + x2 x → √ More elaborately, the function k(x) = sin 1 + x2 can be thought of as the result of composing three functions in series √ √ x → f → 1 + x2 → g → 1 + x2 → h → sin 1 + x2 √ f (x) = 1 + x2 g(x) = x h(x) = sin x The Chain Rule is a rule for differentiating functions that are built up in this way. Let me start with the simple case of two functions. Rule 7: Chain Rule If f and g are differentiable functions and h(x) = f (g(x)) then h0 (x) = f 0 (g(x))g 0(x)

18

CHAPTER 1. THE DIFFERENTIAL CALCULUS Another way to say this is as follows: if y = h(x) then y = f (u) where u = g(x) and dy du dy = dx du dx

This is an easy formula to remember because of the ‘cancellation’. More generally, if we have a whole string of functions composed together, e.g. y = f (u) where u = g(v), where v = h(w), where w = k(x) (so y = f (g(h(k(x))))) then dy du dv dw dy = dx du dv dw dx —you can see the pattern. Example 1.7. f (x) = sin(x2 ) Think of this as y = sin u where u = x2 . Then dy = cos u and du

du = 2x dx

So the Chain Rule says that dy du dy = = cos(u).2x = 2x cos(x2 ) dx du dx (A small note in passing: sin xn always means sin(xn ), whereas (sin x)n is usually written as sinn x.) √ Example 1.8. f (x) = 1 + x2 √ Think of this as y = u, where u = 1 + x2 . Then 1 1 dy = √ du 2 u

and

du = 2x dx

So the Chain Rule says that dy du 1 1 x dy = = √ .2x = √ dx du dx 2 u 1 + x2 Example 1.9. f (x) = sin(1 + cos x) Think of this as y = sin u, where u = 1 + cos x. Then dy = cos u and du

du = − sin x dx

So the Chain Rule says that dy du dy = = cos(u).(− sin x) = − sin(x) cos(1 + cos x) dx du dx

1.9. THE CHAIN RULE

19

q Example 1.10. f (x) = 1 + sin2 (1 + x2 ) This goes deeper. Think of it as y=

√

u = 1 + v2,

u,

v = sin w,

w = 1 + x2

Then 1 1 dy = √ du 2 u

du = 2v dv

dv = cos w dw

dw = 2x dx

So the Chain Rule says that dy dy du dv dw 1 1 2x sin(1 + x2 ) cos(1 + x2 ) p = = √ .2v. cos(w).2x = dx du dv dw dx 2 u 1 + sin2 (1 + x2 ) Note that there is often nothing clear cut about the way in which you choose to break down a function. The usual idea is that you should try to break the function up into pieces that are small enough to differentiate easily. The more experienced you are, the bigger the pieces can become. Example 1.11. f (x) = x2 + tan(x2 ) This requires a combination of rules. First treat it as a sum: d df = 2x + tan(x2 ) dx dx Now do the remaining derivative by chain rule: y = tan u, u = x2 dy du dy = = sec2 (u).2x = 2x sec2 (x2 ) dx du dx So we finally get f 0 (x) = 2x + 2x sec2 x2 x 1 + x2 Here we have to start by using the quotient rule:

Example 1.12.

f (x) = √

√ √ 2 − x. d ( 1 + x2 ) 1. 1 + x dx f 0 (x) = 1 + x2 Now we complete the job by using the chain rule to show that x d √ ( 1 + x2 ) = √ dx 1 + x2 I leave it as an exercise to show that the final result simplifies down to f 0 (x) = 1/(1+x2)3/2 .

20

CHAPTER 1. THE DIFFERENTIAL CALCULUS

Questions 3

(Hints and solutions start on page 66.)

Q 1.18. If f (x) = x2 + 1 and g(x) = 1/x what are f (g(x)), g(f (x)) and g(g(x))? Q 1.19. If f (x) = x + 1, g(x) = 1/x and h(x) = 1 − x what are the functions f (g(h(x))), f (h(g(x))), g(h(f (x))) and so on through all six possibilities? Q 1.20. Differentiate the following, using the chain rule. √ 1 1 + x2 , (2x + 1)5 , (3x − 2) 3 , sin3 x, cos10 x, tan(1 + x2 ),

sin(cos x),

2 sin3 x − 3 cos3 x,

√

2

(x3 + 2x − 6) 3 ,

sin x − cos x,

sin 5x,

1 cos(t + ), t

√ sin2 ( t),

sin

√

sin t,

cos(3x + 1), √

tan x, 1

sin 3 x2 .

Q 1.21. If f (x) = x2 + c what are f (f (x)) and f (f (f (x)))? Q 1.22. Now for a general pile of functions using all the rules. r 1 x x2 + 1 1 1+x √ , , , , x + , 1−x sin(x2 ) x2 − 1 1 + sin4 t 1 + x2 p

2

2

sin x + cos x,

2  √ √ u 1+x 3 tan(tan x), p , sin( 1 + x2 + x), √ , 1−x u− u p sin(2x) sin(3x), 3 sin2 x + 2 cos2 x. u+

Q 1.23. You should know from school that, if x 6= 1, 1 + x + x2 + x3 + x4 + · · · + xn =

xn+1 − 1 . x−1

Differentiate both sides of this equation and put x = −1. Show that you get 1−2+3−4+···±n =

1 − (2n + 1)(−1)n . 4

Check that this is correct for n = 4 and n = 5. *Q 1.24. If f (x) = sin(sin(sin(sin x))) what is f 0 (x)?

1.9. THE CHAIN RULE

21

*Q 1.25. [Only attempt this if you already know about the natural logarithm function ln(x) and that its derivative is 1/x.] Consider the two functions f (x) = ln(x) and g(x) = sin(x). What is the function p(x) = f (f (g(x)))? What is p0 (x)? For what values of x is p(x) defined? For what values of x is p0 (x) defined ?!!? *Q 1.26. s(x) and c(x) are two differentiable functions of x which are defined for all values of x. Suppose that s(0) = 0, c(0) = 1 and that, for all values of x, s0 (x) = c(x) and c0 (x) = −s(x). Let A(x) = s(x)2 + c(x)2 . Show that A0 (x) = 0 and deduce that s(x)2 + c(x)2 = 1 for all values of x. Let B(x) = (s(x) − sin(x))2 + (c(x) − cos(x))2 . Show that B 0 (x) = 0 and deduce that s(x) = sin(x)

c(x) = cos(x)

for all values of x. *Q 1.27. You know that the functions sin x and cos x take values between −1 and 1. You also know that a function with a positive derivative is increasing in value. Consider the function f (x) = x − sin x. Obviously, f (0) = 0. Now differentiate f (x) and show that for 0 < x < π this derivative is positive. Deduce that for this range of values x > sin x (why did I stop at π, and does it really matter?) Use a similar argument to show that, on this range of values, 1 cos x > 1 − x2 2 and

1 x − x3 < sin x < x 6

and, if you still have the energy, 1 1 1 1 − x2 < cos x < 1 − x2 + x4 2 2 24 How accurately does this tell you the value of cos 0.2 and cos 0.1 ? If you are feeling totally masochistic you can try to show that 1 1 1 6 1 1 x < cos x < 1 − x2 + x4 1 − x2 + x4 − 2 24 720 2 24 How accurately do you know cos 0.2 now? What about cos 0.4?

22

1.10

CHAPTER 1. THE DIFFERENTIAL CALCULUS

Proofs of some of the Rules

I cannot give you strict proofs of the rules because they are based on the manipulation of limits and we have not done the theory of limits precisely—I have taken them as being ‘obvious’. But I can indicate the ideas behind the proofs. I will just do two of them, the Sum Rule and the Product Rule. First the Sum Rule. Let f (x) and g(x) be differentiable functions. Let p(x) = f (x) + g(x). By definition, p(x + h) − p(x) p0 (x) = lim h→0 h (f (x + h) + g(x + h)) − (f (x) + g(x)) = lim h→0 h   f (x + h) − f (x) g(x + h) − g(x) + = lim h→0 h h f (x + h) − f (x) g(x + h) − g(x) = lim + lim h→0 h→0 h h 0 0 = f (x) + g (x) as required. You can now easily supply a similar proof for the subtraction rule, it just involves changing a few signs. The Product Rule is a lot more tricky (in fact it managed to fox Newton himself and the rule was first found by Leibniz). The method is similar to the one we have just used, but we have to be rather more canny in how we do the manipulation. Let f and g be as before and let p(x) = f (x)g(x). Now write down the Newton Quotient for p: f (x + h)g(x + h) − f (x)g(x) p(x + h) − p(x) = h h At first sight the numerator of the RHS does not seem to break apart very easily into an f bit and a g bit. But then we remember that the product rule itself (f g)0 = f 0 g + f g 0 is rather complicated. The trick is to write the RHS of the above equation as 1 (f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x)) h All we have done here is add in a term and then subtract it off again. That, as it turns out, is the required sleight of hand. We can now re-bracket this expression as g(x + h) − g(x) f (x + h) − f (x) · g(x + h) + f (x) · h h (check that I have not changed anything!). Now take the limit as h → 0. The two ‘newton quotients’ in the expression tend to the corresponding derivatives in the limit and the only other thing that is changing is g(x + h) which just tends to g(x). So, as h → 0 we get the limit p0 (x) = f 0 (x).g(x) + f (x).g 0(x) and we have proved the Product Rule.

1.11. HIGHER DERIVATIVES

1.11

23

Higher Derivatives

The derivative of a function is a function and it may be possible to differentiate it again— and again and again. The derivative of the derivative is called the Second Derivative   d df d2 f = dx2 dx dx also denoted by f 00 (x).

  d d2 f d3 f also denoted = The derivative of this is called the Third Derivative dx3 dx dx2 by f 000 (x). And so on. n The nth derivative is denoted by ddxnf . In ‘dash’ notation the derivatives are called f 0,

f 00 ,

f 000 ,

f iv ,

f v,

. . . f (n)

So, for example, the tenth derivative of f is written f (10) . Just where the Roman numerals stop and the brackets take over is a matter of personal preference. Example 1.13. If f (x) = x3 + 2x2 + 3x + 1 then f 0 (x) = 3x2 + 4x + 3 f 00 (x) = 6x + 4 f 000 (x) = 6 f iv (x) = 0 f v (x) = 0 and so on. If f (x) = sin x then f 0 (x) = cos x f 00 (x) = − sin x f 000 (x) = − cos x f iv (x) = sin x f v (x) = cos x and so on.

1.11.1

Geometrical Interpretation of Second Derivative

f 0 (x) gives the slope of the function—the ‘rate’ at which f is changing. f 00 (x) gives the rate at which the slope is changing and gives, roughly, the rate at which the graph is bending.

24

CHAPTER 1. THE DIFFERENTIAL CALCULUS

f’’(x) < 0

f’’(x) > 0

Figure 1.5: The sign of the second derivative shows the shape of the curve. You will often hear politicians, in desperation, talking about the rate of decrease of the rate of increase of something unpopular. Look at the graphs in Fig 1.5 In the first the slope is increasing as x increases, so the derivative of the slope is positive. f 00 (x) > 0 (Remember that slope is a signed quantity. Most people would say that the first graph slopes more at the left-hand end than it does in the middle and that therefore the slope is initially decreasing rather than increasing. Actually, it is going from very negative to less negative and that is an increase—(-2) is bigger than (-3).) In the second the slope is decreasing as x increases, so the derivative of the slope is negative. f 00 (x) < 0 Example 1.14. Fig 1.6 indicates the various possibilities

f’(x) > 0

f’(x) < 0

f"(x) < 0

f’(x) > 0

f"(x) > 0

Figure 1.6: First and second derivatives geometrically.

1.11.2

Kinematical Interpretation

If a particle is moving along the x-axis so that its position at time t is given by x(t), then we have seen that x(t) ˙ is the velocity of the particle at time t. The second derivative, denoted by x ¨(t), gives the rate of change of velocity—the acceleration.

1.11. HIGHER DERIVATIVES

25

This is a very important concept because Newton’s Laws of motion relate the force acting on a particle to its acceleration. Example 1.15. Suppose that a particle has position x(t) = A sin(ωt) at time t, where A and ω are constants. Find its velocity and acceleration. Its velocity is x(t) ˙ = Aω cos(ωt) and its acceleration is x¨(t) = −Aω 2 sin(ωt).

Questions 4

(Hints and solutions start on page 68.)

Q 1.28. Work out the first ten derivatives of f (x) = sin x. What is the 300th derivative of f (x)? Now work out the first ten derivatives of f (x) = x sin x. Can you guess the value of f (300) (x)? √ Q 1.29. Work out the first 5 derivatives of f (x) = x. Can you work out an expression for the nth derivative of f (x) using factorials? (Last bit is quite hard.) Q 1.30. Show that the function x(t) = A sin(ωt + α), where A, ω and α are constants, satisfies the so-called Harmonic Oscillator equation x¨ = −ω 2 x Q 1.31. Show that y = (cos(2x) + sin(2x))/x satisfies the equation 2 y 00 + y 0 + 4y = 0 x *Q 1.32. I have a function f (x). I differentiate it repeatedly and eventually end up getting zero. What kind of function must f have been? *Q 1.33. The product rule gives you a formula for (f g)0. Work out similar formulas for the second, third and fourth derivatives of f g. If you know about such things as Binomial Coefficients, or the Pascal triangle, you can guess at the formula for the nth derivative (Leibnitz’s Theorem). *Q 1.34. Let T stand for tan(x) and let Tk stand for the k th derivative of tan(x). Why is T1 = 1 + T 2? Use the chain rule to show that T2 = 2T + 2T 3 and T3 = 2 + 8T 2 + 6T 4. Also work out T4 , . . . , T7 . Now look at the pattern of the coefficients of all these polynomials (make up a table). Confirm the following claims in the cases that you have calculated: (1) Tn is a polynomial of degree n + 1 in T ; (2) if n is odd then Tn just involves even powers of T , and vice versa if n is even; (3) the coefficient of T n+1 in Tn is n!. Can you spot a formula for the second highest coefficient of Tn ? Can you now have a shot at justifying these claims in the general case — not just in the examples that you have worked out.

26

CHAPTER 1. THE DIFFERENTIAL CALCULUS

n  √ *Q 1.35. If y = x + 1 + x2 show that (1 + x2 )y 00 + xy 0 − n2 y = 0. (Could try writing y = z n .) *Q 1.36. Let fn (x), where n is a positive integer, be given by f (x) = xn if x ≥ 0 and f (x) = 0 if x < 0. Sketch the graphs of f1 , f2 and f3 . Show that f1 (x) is not differentiable at x = 0. What is the derivative of f2 (x)? How many times can we differentiate f2 (x)? How many times can we differentiate fn (x)?

1.12

Limits & Continuity

This is a brief section meant to give you the basic idea about these things. A proper treatment will have to wait until second year.

1.12.1

Limits

I have been using the idea of a ‘limiting value’ of a function, limx→a f (x), and I have been taking the meaning as being ‘obvious’—though it really ought to be defined properly (see later year). Of course, limits may not exist. For example, we would say that x1 does not have a limit as x → 0. In this kind of case we often write something like 1 → ∞ as x

x→0

More complicatedly the function f (x) = sin x1 does not have any kind of limiting value as x → 0. It is reasonably easy to calculate with limits, if we know that they exist and behave themselves. We have the following simple rules for some simple situations. If f (x) and g(x) are functions that have limiting values α and β as x → a: lim f (x) = α

x→a

lim g(x) = β

x→a

then f (x) ± g(x) → α ± β f (x)g(x) → αβ f (x)/g(x) → α/β if

Questions 5

β 6= 0

(Hints and solutions start on page 69.)

Q 1.37. What is the limiting value of f (x) = 1/x as x → ∞? What is the limiting value of f (x) = cos(1/x) as x → ∞?

1.12. LIMITS & CONTINUITY

27

Q 1.38. Work out the values of the following limits x2 + 1 , x→∞ x2 − 1

1 , x→∞ 1 + x2 lim

lim

lim

√

x→∞

x+1−

√  x .

I suggest that you try dividing top and bottom by x2 in the second one. For the last one try a − b = (a2 − b2 )/(a + b). Q 1.39. Consider the function f (x) = (x2 − 1)/(x − 1). If we put x = 1 in the formula we get f (0) = 0/0, which makes no kind of sense. What is the limiting value of f (x) as x approaches 1? (This is where you tend to hit a doctrinal difference between Pure and Applied mathematicians. Pure mathematicians tend to say that the function f (x) is not defined for x = 1 though, as you should have seen, it has a perfectly decent limiting value as x → 1. Applied mathematicians tend to argue that f (x) = ((x − 1)(x + 1))/(x − 1) = x + 1 and is therefore a perfectly well defined function everywhere.)

1.12.1

Continuity

Funny business with quotients apart, you would normally expect to be able to work out limx→a f (x) simply by putting x = a in f (x). In other words, you would expect that as

x→a

f (x) → f (a)

This may not be true. For example, consider the function given by ( 1 x≥0 f (x) = 0 x 0 between a and b (a < b). Then (b − a)f 0 (c) must be positive (whatever c is). So f (b) must be bigger than f (a). So, in an obvious sense, the function must be increasing. Here’s another typical application. Let f (x) = tan(x). Then f 0 (x) = 1/ cos2 (x) > 1. So we must have tan(b) − tan(a) > b − a This result is rubbish. For example, tan(π) = tan(0) = 0. But 0 − 0 is not bigger than π − 0. What’s gone wrong? The problem here is that tan(x) is not a differentiable function between 0 and π—it blows up at π/2. When using these theorems, no matter how ‘obvious’ they are, you really must check that the conditions of the theorem are precisely satisfied.

Chapter 2 Further Differentiation 2.1 2.1.1

The Inverse Trigonometric Functions Inverse Functions in general

If y = 3x − 1 then x = (y + 1)/3. The first formula defines a function f (x) = 3x − 1 x → f → 3x − 1 The second formula gives the inverse (reverse) of this function, which we usually denote by the very unfortunate notation f −1 (x). x → f −1 →

x+1 3

All that we are doing is switching round the input and output of the machine—putting in at the ‘out’ end and getting out at the ‘in’ end. The inverse function undoes the effect of the original function: x → f → f −1 → x (Note the notational problem. Does f −1 mean the inverse function of f or does it mean 1/f ? I’m afraid that the answer has to be ‘it depends on the context’. You hope that it is obvious which interpretation the author has in mind. If it is not, then find another author! We really should have a better notation but once these things settle down it is very difficult to change them.) There are problems. We cannot always define an inverse in a straightforward way. Consider the function f (x) = x2 . What can we make of f −1 ? If we put x = 1 into the f machine we get out 1. If we put −1 into the machine we also get out 1. So what do we expect to get out of the inverse machine if we put in 1? You see the problem. This machine is not uniquely reversible. 1 → f −1 →????

1, −1 → f → 1 31

32

CHAPTER 2. FURTHER DIFFERENTIATION

The problem that we face here is described by the technical term one-to-one. A function f is one-to-one if you never have f (x) = f (y) when x 6= y. The function f (x) = x2 is NOT one-to-one because, for √ example, f (1) = f (−1). Nevertheless, we can say that, in some sense, g(x) = x is an inverse for f . Certainly, most people would say that g undoes the effect of f x → f → x2 → g → x but this is not quite correct as it stands. It becomes correct if we put on the extra condition that x is not allowed to be negative and insist that the result of the square root is not negative.

Questions 6

(Hints and solutions start on page 69.)

Q 2.1. What are the inverses of the functions f (x) = 2x + 4 and g(x) = 2 − 5x? Q 2.2. What is the inverse of the function f (x) = 1/x for x > 0? Q 2.3. The function f (x) is defined as follows: if x ≥ 0 then f (x) = 1 + x and if x < 0 then f (x) = 1 + x/2. What is the inverse of f ? *Q 2.4. If the inverse of the function f (x) for some range of values of x is also f (x) what does that tell you about the graph of f (x) on that range? If f (x) is a differentiable function and f (f (x)) = x for all values of x can f be anything other than ±x? (This is a problem for thinking about rather than answering precisely.) *Q 2.5. Suppose that a, b, c, d are constants. Consider the following function (known as a M¨obius Transformation) ax + b f (x) = . cx + d Suppose that ad 6= bc. Show that if c 6= 0 there is one value of x at which f (x) is not defined and that there is one number that is not a possible value of f . Show that if you forget about the point where f is not defined then f is a one-to-one function. What is the inverse of f (if you exclude the bad points)? Show that any M¨obius Transformation can be built up as a composition of some number of the functions p(x) = x + u, r(x) = vx (v 6= 0) and q(x) = 1/x (note that these are all M¨obius Transformations themselves).

2.1.1

The Inverse Trigonometric Functions

The Inverse Trigonometric Functions are meant to be the ‘inverses’ of the standard trigonometric functions. There are obviously going to be difficulties in trying to define them because none of the standard trig functions are one-to-one.

2.1. THE INVERSE TRIGONOMETRIC FUNCTIONS

33

We have to start somewhere so let me start by giving the ‘obvious’ (and obviously wrong) definitions of the inverse trig functions. I will then try to put them right. The inverse functions of sin, cos and tan are called arcsin, arccos and arctan. The notations sin−1 (x), cos−1 (x) and tan−1 (x) are also common, though they tend to be confusing to beginners. arcsin(x) is the angle whose sine is x arccos(x) is the angle whose cosine is x arctan(x) is the angle whose tan is x if y = sin x then x = arcsin y if y = cos x then x = arccos y if y = tan x then x = arctan y That’s what we want to mean by the inverse trig functions, but the definitions do not make sense as written, simply because the trigonometric functions are not one-to-one. For example, sin 0 = 0 and sin π = 0 and so on. So what is arcsin 0? Is it 0 or π or what? We get round this difficulty by carefully restricting the values that the inverse trigonometric functions can take. Let me start with sin x. This function is not one-to-one, but it becomes one-to-one if we re1 strict attention to the range −π/2 ≤ x ≤ π/2. It also takes all its possible values in this range. If we make the range any wider then the function ceases to be one-to-one. If we make the range any 0 π narrower then the function does not take all its − π2 0 2 possible values. Definition 2.6. arcsin(x) is the angle between − π2 and π2 whose sine is x. (−1 ≤ x ≤ 1) For the cosine we have to pick a different range because cosine is not one-to-one on the range −π/2 ≤ x ≤ π/2. Instead, we choose the range 0 ≤ x ≤ π. This works.

-1

Figure 2.1: Graph of sin x.

Definition 2.7. arccos(x) is the angle between 0 and π whose cosine is x. (−1 ≤ x ≤ 1) The graphs of the inverse trigonometric functions are given in Figs. 2.2 and Fig. 2.3. These show the general pattern for the graphs of inverse functions. All we are doing is reversing the original x and y axes — which is the same thing as flipping the graph over the line y = x. The resulting graph is a genuine graph (doesn’t give more than one y value for any given x value) provided that the original function has an inverse.

For the tan we can use more or less the same range as we did for sine, except that we have to leave out the end points −π/2 and π/2 because tan is not defined at these values (it blows up).

34

CHAPTER 2. FURTHER DIFFERENTIATION π 2

π y

y

arcsin(x)

arccos(x) x

-1

π 2

1

x

− π2

-1

Figure 2.2: Graph of arcsin(x).

1

Figure 2.3: Graph of arccos(x).

Definition 2.8. arctan(x) is the angle in the range −

π π 0 and θ is the angle from OX round to OP antiFigure 2.4: Polar co-ordinates clockwise. By convention, we take the angle range to be −π < (r, θ) of a point. θ ≤ π. (some people take 0 ≤ θ < 2π). Note that we do not give polar coordinates for the origin, because the angle does not make sense there. Now consider the problem of converting between cartesian and polar coordinates. One way round is easy: x = r cos θ y = r sin θ (these are really just the definitions of sin θ and cos θ.) The other way round is more tricky and needs some understanding of inverse trig functions. First of all, without any difficulty, p r = x2 + y 2 The real problem is to get θ in terms of x and y. Dividing the above equations gives tan θ = So we are tempted to write

y x

y θ = arctan( ) x

This is wrong. First of all, we have to be careful about the case x = 0 (which corresponds to the y-axis) because we don’t want to divide by zero. Inspection of the picture tells us that if x = 0 then if y > 0 then θ = π/2 if y < 0 then θ = −π/2 Our problems don’t end there. Recall that arctan is defined as taking values in the range π π − 0 then θ = arctan(y/x) if x < 0 then if y ≥ 0 then θ = arctan(y/x) + π if y < 0 then θ = arctan(y/x) − π

Questions 7

(Hints and solutions start on page 69.)

Q 2.9. Get used to the inverse trig buttons on your calculator by checking the following arcsin 0.7 ≈ 0.7754

arccos(0.1) + arcsin(0.1) ≈ 1.5708

tan−1 0.7 ≈ 0.6107

Q 2.10. Give the values of the following without using your calculator (check afterwards). Remember what you know about values of the trig functions! arcsin(1),

arcsin(−1),

arctan(0),

cos−1 (1),

arctan(1),

arccos(−1), arccos(0) √ sin−1 (1/ 2), arccos(1/2)

Q 2.11. Without using a calculator, find the polar coordinates of the following points: (1, 0), (0, 1), (−1, 0), (0, −1), (1, 1), (−1, 1), (−1, −1). Now use your calculator to find the polar coordinates of the following points: (2, 1), (−3, 4), (5, −2). Q 2.12. Differentiate the following functions. sin−1 (2x), x. tan−1 x, √ √ tan−1 (1 + x2 ), arcsin( 1 − x2 ), cos−1 x

2 arcsin x − 3 arccos x, arccos(3x + 1),

Q 2.13. Differentiate the function f (x) = arctan(x) + arctan(1/x). Explain the answer that you get (think in terms of a right-angled triangle). *Q 2.14. Sketch the graph of the function f (x) = arcsin(sin x) for 0 ≤ x ≤ 3π. Do the same for g(x) = arccos(cos x) and h(x) = arcsin(cos x). *Q 2.15. Show that if a and b are any constants then y = (arcsin x)2 + a arcsin(x) + b satisfies the equation (1 − x2 )y 00 − xy 0 = 2.

38

CHAPTER 2. FURTHER DIFFERENTIATION

*Q 2.16. This is nothing to do with Inverse Trig functions. You know that tan 2x =

2 tan x 1 − tan2 x

It can be shown that the approximation tan x ≈ x+ 13 x3 gives an error of less than 1.4×10−11 if |x| < 0.01. Use both of these results to obtain an approximate value for tan 0.1 by a process of repeated halving and doubling (I mean you to use a calculator). Show that if you know the value of tan x, with 0 ≤ x ≤ π/2 then you can easily work out the values of sin x and cos x, so long as you can take square roots. Get an approximation to sin 0.1. By inspecting the graphs of the trig functions show that you can work out the sin and cos of any angle once you know the values of sin and cos for angles in the range 0 ≤ x ≤ π/2 (at least, in principle). Put all this together to calculate sin 6.4 to an accuracy of 5 decimal places. You can check your answer with your calculator. (If you know anything about computer programming you could now write a decent program to evaluate the trig functions.) Why did I say “in principle” a couple of paragraphs ago?

2.2

Implicit Functions and their Differentiation

Suppose that the variable y is a function of the variable x. Up to now the definition of y in terms of x has been given in the form y = expression in x This is called an Explicit Formula for y in terms of x. You get the value of y by plugging the value of x into the RHS. It is also possible, with a bit of care, to define y in terms of x by a equation of more general form involving x and y. For example, the equation of a straight line: ax + by + c = 0 or the equation of a circle: or, more elaborately,

x2 + y 2 = r 2 x2 y 2 + y 4 = sin(xy)

Such an equation is called an Implicit Formula for y in terms of x. In some cases it is easy to turn an Implicit Equation into an Explicit Equation, simply by ‘solving for y’. For example, if b 6= 0 in the equation of the straight line, then the equation can be changed to c a y =− x− b b

2.2. IMPLICIT FUNCTIONS AND THEIR DIFFERENTIATION

39

which is now explicit. As another example, the Implicit Equation 3y + x2 = 2x2 y can be rearranged to give the Explicit Equation x2 y= 2 2x − 3 Things are not always so simple. There are two problems that we face. The first and most obvious is that we may not know how to solve the equation for y — see the third example at the start, you cannot do it and nor can I. The second problem is more theoretical. An implicit equation on its own may not define y as a function of x in a straightforward way. Consider the example of the circle equation x2 + y 2 = r 2 . If we ‘solve for y’ we actually get two explicit equations: √ √ y = r 2 − x2 andy = − r 2 − x2 Since, within the range −r < x < r there are always two y-values for each x-value we cannot hope to express y as a single function of x. So when we talk of the function defined by an Implicit Equation we should always be aware that there may be more than one and that we may need extra information to decide on which to choose.

2.2.1

Differentiating Implicitly Defined Functions

Suppose that y is a function of x that satisfies the Implicit Equation x2 + y 2 = sin(x + y) Can we work out the derivative dy/dx? This would be no problem if we could rearrange our equation into an Explicit equation y = . . . , but in this case we cannot. Even if we cannot convert the equation into explicit form it is still quite easy to work out the derivative by using the chain rule. Just differentiate both sides of the equation, remembering that y is a function of x and that we will have to use the Chain Rule on it. One or two examples will show you how easy this is. Example 2.3. x2 + y 2 = 1 Differentiating both sides wrt x we get 2x +

d 2 (y ) = 0 dx

By the chain rule d dy d 2 (y ) = (y 2). dx dy dx So the equation becomes x dy =− dx y Note the almost inevitable snag: the expression for the derivative contains both x and 2x + 2yy 0 = 0 or

y.

40

CHAPTER 2. FURTHER DIFFERENTIATION

Example 2.4. Find the equation of the tangent line to the curve x2 + y 3 = 2 at the point (1, 1) on the curve. To find the equation of the tangent we need to find the slope of the curve at (1, 1). This means working out the derivative. We differentiate both sides of the equation with respect to x. The derivative of x2 is 2x and, by the chain rule, the derivative of y 3 is 3y 2 y 0. So we get 2x 2x + 3y 2y 0 = 0 hence y 0 = − 2 3y So the slope of the curve at the required point is y 0 = −2/3. So the equation of the tangent 2 is y − 1 = − (x − 1). 3 Example 2.5. x3 + sin(xy) = xy 2 Differentiating both sides wrt x we get 3x2 +

d d sin(xy) = 1.y 2 + x y 2 dx dx

This still leaves us with some work to do: d 2 y = 2yy 0 dx as before. The sin term involves a double application of the chain rule. We have, let’s say, z = sin u where u = xy. So dz du d dz = = cos(u) (xy) dx du dx dx Now d d (xy) = 1.y + x. y = y + xy 0 dx dx Finally we get 3x2 + (y + xy 0) cos(xy) = y 2 + 2xyy 0 We can now rearrange this to get all the y 0 terms onto the LHS: (x cos(xy) − 2xy)y 0 = y 2 − 3x2 − y cos(xy) So

y 2 − 3x2 − y cos(xy) dy = dx x cos(xy) − 2xy

—a mess, but a successful mess. Example 2.6. Find the tangent at (0, 0) to the curve (x2 + y 2)2 = x2 − y 2. Differentiate implicitly to get 2(x2 + y 2)(2x + 2yy 0) = 2x − 2yy 0ory 0 =

x(1 − 2x2 − 2y 2) y(1 + 2x2 + 2y 2 )

2.2. IMPLICIT FUNCTIONS AND THEIR DIFFERENTIATION

41

Now put x = 0 and y = 0 — and get problems: 0/0. It doesn’t work. What’s happening? We are interested in the point (0, 0). Suppose that we are very close to it, so that x and y are very small. Then x2 and y 2 are extremely small and (x2 + y 2 )2 is absolutely tiny. So, rather approximately, we can say that, near (0, 0), the equation is x2 − y 2 ≈ 0 or y 2 ≈ x2 or y ≈ ±x. This gives two straight lines through the origin at right angles—which explains why we had difficulty in finding the slope at that point. In fact the curve, a Lemniscate, is shown in Fig. 2.5

Figure 2.5: The lemniscate of Example 2.6. Example 2.7. Here is another kind of example of the use of implicit differentiation. dy We are told that the function y(x) satisfies the equation + xy 2 = x + 1 for all values dx of x. We are also told that y(0) = 1. What, in that case, are y 0(0), y 00 (0) and y 000 (0)? Put x = 0 and y = 1 into the equation and get y 0 (0) + 0 = 0 + 1. So y 0 (0) = 1. Differentiate the equation wrt x and get y 00 + y 2 + 2xyy 0 = 1 Put x = 0 in this, together with y = 1 and y 0 = 1 and get y 00 (0) = 0. Differentiate again wrt x and get y 000 + 4yy 0 + 2xy 02 + 2xyy 00 = 0 Put x = 0 in this together with y = 1, y 0 = 1, y 00 = 0 and get y 000 (0) = −4.

Questions 8

(Hints and solutions start on page 70.)

Q 2.17. For each of the following implicit equations find out the equation of the tangent line at this point

dy dx

at the given point. Also work

x2 + x + y 2 = 1at(0, 1) x3 + y 3 = 9at(2, 1) x3 + xy + y 3 = 1at(1, 0) π π sin(x + y) + cos(x − y) = 0at( , − ) 4 4 (x2 + y 2 )2 + (x2 − y 2)2 = 4at(1, 1)

42

CHAPTER 2. FURTHER DIFFERENTIATION

*Q 2.18. Show that the equation of the tangent to the curve x2 y 2 + 2 =1 a2 b at the point (x0 , y0) on the curve can be written as xx0 yy0 + 2 =1 a2 b *Q 2.19. The problem in this question is to work out the shape of the curve produced by all the points P such that the product of the distances from P to two given points A and B is a constant: P A.P B = α. Suppose we choose axes so that A = (−p, 0) and B = (p, 0). Show that the equation of the curve can be written as (x2 + y 2 + p2 )2 = α2 + 4p2 x2 . Note that the curve must be symmetrical about both the x and the y axes (why?). Find where, if anywhere, the curve cuts the axes. Now try to find the rough shape of the curve. The shape depends on the value of α. What is special about the case α = p2 ?

2.3

Parametric Representation

Suppose we have a point moving around in the (x, y) plane. At each time t the particle will be at some point whose coordinates we can write as (x(t), y(t)). i.e. the x and y coordinates of the point are given as functions of the parameter t. Many curves can be most conveniently expressed in the form x = x(t)

y = y(t)

where t is some parameter. t need not be time, though it often helps to think about it in that way. This is called a parametric representation of the curve. It may in some cases be possible to eliminate t between the two equations and get an Implicit Equation just involving x and y. Even when possible, this is not always desirable. Example 2.8. The Circle Any point on the circle x2 + y 2 = r 2 can be written in the form x(θ) = r cos θ y(θ) = r sin θ — just using polar coordinates. This is now a parametric representation of this circle. As θ goes from 0 to 2π the point (x(θ), y(θ)) goes once round the circle.

2.3. PARAMETRIC REPRESENTATION

43

Similarly, the circle (x − a)2 + (y − b)2 = r 2 can be represented parametrically as x(θ) = a + r cos θ y(θ) = b + r sin θ There are always lots of different ways of parameterizing a given curve, some more helpful than others. As a trivial example, x(t) = sin(t), y(t) = cos(t) still gives a circle, as does x(t) = cos(t2 ), y(t) = sin(t2 ). Notice that a parametric representation can represent a whole curve in situations where is is not possible to do this by a formula of the form y = f (x), because there may be more than one value of y corresponding to a given value of x. This representation can even handle cases where a curve crosses over itself, as in the case x(t) = cos(t),

y(t) = sin(2t)

which produces a shape like the ∞ sign.

2.3.1

Differentiation in Parametric Form

Suppose that we are given a curve in the parametric form x = x(t)

y = y(t)

where t is a parameter. Can we find the value of dy/dx (in other words, the slope of the curve) without actually having to eliminate t between the equations? The answer is yes and the process is simple and based on the following simple consequence of the chain rule: dy dx dy = · dt dx dt hence, y˙ dy = dx x˙ where the dot denotes differentiation wrt t. Note something that might be a problem in some cases: the answer is given in terms of the parameter t rather than in terms of x and y. Example 2.9. Consider the circle in the parametric form x = r cos t

y = r sin t

Then x˙ = −r sin t and so

y˙ = r cos t

y˙ r cos t dy = = = − cot t dx x˙ −r sin t

44

CHAPTER 2. FURTHER DIFFERENTIATION

Example 2.10. The parabola y 2 = 4ax can be parametrised as x(t) = at2

y(t) = 2at

where t takes all values. Why is this true? Firstly, y 2 = 4a2 t2 and 4ax = 4a2 t2 so the point (x(t), y(t)) certainly satisfies the equation and lies on the parabola. That’s half the story. The second question is: do we get the whole parabola? i.e. is there a value of t corresponding to each point on the curve? This question is easy to answer in this case. Let (x, y) be a point on the parabola, so y 2 = 4ax. Let t = y/2a. Then y = 2at and x = y 2/4a = at2 . Final slight worry: do different values of t give us the same point on the parabola? No they do not — as the above argument actually shows. Put another way: if (at2 , 2at) = (as2 , 2as) then we must have s = t, as can be seen by comparing the y-coordinates. Having settled all that, let’s differentiate: x˙ = 2at

y˙ = 2a so

y˙ 1 dy = = dx x˙ t

To expand on the points raised in the previous example, consider the curve given parametrically by x(t) = at4

y(t) = 2at2

where t takes all values. What curve is this? Well, it is easy to see that x(t) and y(t) satisfy y 2 = 4ax. So we seem to have the parabola again. No we don’t. We get half the parabola. In this parametric representation, assuming that a > 0, the y value can never be negative (2at2 ). Furthermore, we get half the parabola twice over: (at4 , 2at2 ) = (as4 , 2as2 ) if t = ±s. If we think of (x(t), y(t)) as the path of a particle parametrised by time then the particle comes in from infinity along the top half of the parabola, gets to the origin, stops and then reverses back the way it came.

Questions 9

(Hints and solutions start on page 70.)

Q 2.20. For each of the following parametric representations find the parameter. x(t) = t2 ,

dy dx

at the given value of

y(t) = t3 att = 1 π x(t) = cos(t), y(t) = sin(t)att = 4 2 2 x(λ) = λ + 1, y(λ) = λ − 1 at λ = 1 1 y(t) = 2t/(1 + t2 ) at t = x(t) = (1 − t2 )/(1 + t2 ), 2

2.3. PARAMETRIC REPRESENTATION

45

Q 2.21. Consider the curve given parametrically by x(t) = t2

y(t) = t3

Find the slope of the curve at the point with parameter value t. Now try to sketch the curve. Hints: what’s the relation between the points with parameter t and −t?. What exactly is happening at the origin? *Q 2.22. Show that if you put t = tan(θ/2) in the following formulas x(t) =

1 − t2 1 + t2

y(t) =

2t . 1 + t2

(1)

then you get x(θ) = cos θ, y(θ) = sin θ. So (1) seems to be a parametrisation of the circle x2 + y 2 = 1. This is rather interesting because (1) does not use trig functions and does not use square roots either. Is it actually true that you get every point on this circle from (1)? *Q 2.23. Sketch the curve given parametrically by the equations x(t) =

1 1 cos t − cos 2t, 2 4

y(t) =

1 1 sin t − sin 2t 2 4

Be sure to give decent arguments for what you draw. (For those of you who know about such things, this is the main blob of the Mandelbrot Set.) *Q 2.24. L1 and L2 are two straight lines. Initially L1 lies along the y-axis and L2 lies along the line y = 1. The two lines now start to move. L1 turns clockwise about the origin and L2 moves downwards. At time t the line L1 makes angle πt/2 with the y-axis and L2 is the line y = 1 − t. Show that the point of intersection of the two lines moves on the curve ((1 − t) tan(πt/2), 1 − t), called the Quadratrix of Hippias. Try to give a rough sketch of the curve (in the first quadrant). *Q 2.25. This is similar to the previous example. This time the two lines are initially parallel to the y-axis. L1 passes through P = (−1, 0) and L2 passes through Q = (1, 0). L1 rotates around P and L2 rotates around Q, both clockwise. L1 is turning at twice the rate that L2 is turning. Sketch the path followed by the point of intersection of the two lines.

46

CHAPTER 2. FURTHER DIFFERENTIATION

Chapter 3 The Exponential and Logarithm Functions You are already familiar with some of the ‘standard functions’ of mathematics — polynomials, rational functions, square roots, trigonometric functions and their inverses. We are now going to complete the list of standard functions by introducing two more basic functions. These are the exponential function and the logarithm.

3.1

Exponential Population Growth

Let’s start by looking at a problem in biology concerning the growth of animal populations. The theory that I am going to present here is usually associated with Malthus—whose work had an enormous influence in the last century, particularly on Charles Darwin. The theory is not taken very seriously nowadays, but it still makes for a nice illustration. Suppose a certain group of animals or plants has population P (t) at time t. We want to find out something about how this population changes in time. We cannot deduce anything without making some assumptions. I am going to study a very simple ‘growth model’ which starts from the following assumptions. Each year each individual ‘produces’ on average b new individuals. Each year, for each individual in the population, d individuals die. We assume that b and d are constants. In a short time interval from time t to time t + δt we can say, to the first approximation, that b.δt.P (t)individuals are produced d.δt.P (t)individuals die So, with this approximation, P (t + δt) = P (t) + b δt P (t) − d δt P (t) 47

48

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

Rearrange this equation as P (t + δt) − P (t) = (b − d)P (t) δt You will notice that the left-hand side of this equation is just the usual Newton Quotient for P(t). Now let δt → 0 and get dP = κP dt

Exponential Growth Law

where κ = b − d is what is called the growth rate. We have deduced from our assumptions a mathematical relationship based on the time derivative of P . Suppose we start at time t = 0 with population P (0). If κ > 0 then P˙ > 0 so the population is increasing. As it increases the RHS of the equation gets bigger, so the rate of increase also increases—the growth feeds upon itself. The population grows more and more rapidly. If κ = 0 then P˙ = 0 always, so the population remains constant. If κ < 0 then P˙ < 0 at first, so the population is decreasing. As it decreases (so long as it does y not die out) the RHS gets smaller, so the rate of decrease slows down. We will see later that, in fact, the population → 0 as t → ∞. In real life κ>0 this just means that the population dies out. We can give a rough sketch of the three cases κ=0 as shown in Fig. 3.1. In the next section we are goP (0) ing to take a more careful look at the exponential κ 1. Now exp(4) = exp(2) exp(2) and exp(8) = exp(4) exp(4) and so on up. In general, n−1

exp(2n ) = exp(2)2

Since exp(2) > 1 its powers get bigger and bigger without limit. That’s what we wanted. On the other side: exp(−x) = 1/ exp(x) so exp(−x) → 0 as x → ∞ and that proves the second point. That’s the end of my basic work on the exponential function.

3.2.3

Summary

You can open your eyes now. Here is a summary of the information that we have obtained about the exponential function. Even if you did not follow the arguments of the previous section you should still make sure that you know the following information. The exponential function exp(x) (or ex ) is the single solution of dy =y dx

y(0) = 1

It has the properties d x e = ex dx

e0 = 1

e−x =

The number e1 = exp(1) is called e and has the value e = 2.71828 . . . . ex is always positive and increasing. ex → ∞ as x → ∞ ex → 0 as x → −∞ Part of the graph of ex is shown in Fig 3.2.

1 ex

ex+y = ex ey

4

3

2

1

y = ex

3.2. THE EXPONENTIAL FUNCTION

3.2.4

53

Differentiation Examples

These are some more examples of differentiation, now that we have got a new function to differentiate. There is nothing new in the methods. Example 3.1. y = e4x This is a simple, but important, example of the chain rule. We can express the function as y = eu where u = 4x. So, by the chain rule, dy du dy = = 4eu = 4e4x dx du dx In general, if a is a constant, d ax e = aeax dx Example 3.2. y = e3x (sin x + 2 cos x) Think of this firstly as a product d 3x dy = (e )(sin x + 2 cos x) + e3x (cos x − 2 sin x) dx dx = 3e3x (sin x + 2 cos x) + e3x (cos x − 2 sin x) = e3x (sin x + 7 cos x) 2

Example 3.3. f (x) = ex Chain rule again. Let y = eu where u = x2 . Then dy du dy 2 = = eu .2x = 2xex dx du dx Example 3.4. Show that the function y = ex (sin(x) + cos(x)) satisfies the equation y 00 − 2y 0 + 2y = 0 for all values of x. The basic plan is straightforward. Work out y 0 and y 00 , substitute them into the equation and hope for the best. This is a case where the differentiation is quite easy but is also quite messy. It is important to keep workings clear and well-organized or you are bound to make a silly mistake somewhere. y 0 = ex (sin(x) + cos(x)) + ex (cos(x) − sin(x)) It is best to tidy this up at once to get y 0 = 2ex cos(x) Now differentiate again.

y 00 = 2ex cos(x) − 2ex sin(x)

54 So

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

y 00 − 2y 0 + 2y = 2ex cos(x) − 2ex sin(x) − 4ex cos(x) + 2ex (sin(x) + cos(x))

If you now simplify down the RHS of this equation you will find that you get the answer 0, regardless of the value of x. That solves the problem. Example 3.5. Find the values of α for which y = eαx sin(2x) satisfies the equation y 00 − 4y 0 + 8y = 0 This is much the same as the previous example. We calculate y 0 and y 00 and then put the results into the equation. We then hope that the fact that the equation has to be satisfied for all values of x will tell us the possibilities for α. Again, good organization is important—simplify as you go along, or things get out of hand. y 0 = αeαx sin(2x) + 2eαx cos(2x) = eαx (α sin(2x) + 2 cos(2x)) y 00 = αeαx (α sin(2x) + 2 cos(2x)) + eαx (2α cos(2x) − 4 sin(2x)) = eαx ((α2 − 4) sin(2x) + 4α cos(2x)) So y 00 − 4y 0 + 8y = eαx ((α2 − 4) sin(2x) + 4α cos(2x) − 4α sin(2x) − 8 cos(2x) + 8 sin(2x)) = eαx ((α2 − 4α + 4) sin(2x) + (4α − 8) cos(2x)) and we want to know what values of α make this identically zero. We can forget about the exponential at the start (that will never be zero). The expression will only vanish identically if the coefficients of the sine and the cosine both vanish (see exercises for the justification for this). The cosine term tells us that the only possibility for α is 2 and, if you check, you will see that α = 2 kills off the sine term as well. So the answer to the question is that α has to be 2.

3.3

The Natural Logarithm

The Logarithm is an older function than the Exponential. It is also seen as being more elementary—in the sense that you are likely to meet logarithms at a fairly early stage of your mathematical education (at least, you did before the days of pocket calculators). The main elementary use of the logarithm is that it can be used to turn multiplication into addition: log(ab) = log(a) + log(b), thus making calculations easier to do. That aspect of the logarithm has rather faded away with the advent of pocket calculators (life was hard when I was young). Most calculus books still tend to take the logarithm as basic and derive the exponential function from it. I am adopting the opposite approach of starting with the exponential

3.3. THE NATURAL LOGARITHM

55

function (which really is more important) and then deriving the logarithm function from it. You may have studied logarithms ‘to base 10’ in school. You can now forget about them.

3.3.1

Definition of the Logarithm

We have seen in the previous section that the exponential function is positive and always increasing. Consequently, given any positive value y there is one and only one value x for which exp x = y. This means that we can construct an inverse function for exp. The inverse of the exponential function is called the Natural Logarithm and is denoted by ln x. As immediate consequences of the definition we have: ln(exp(x)) = ln(ex ) = xfor all x exp(ln(x)) = eln x = xfor x > 0 Reading some of the properties of the exponential ‘backwards’ we get: ln(x) is defined for x > 0 but not for x ≤ 0. ln(1) = 0, ln(e) = 1, ln(x) < 0 for 0 < x < 1 and ln(x) > 0 for x > 1. ln(x) is an increasing function and ln(x) → ∞ as x → ∞, ln(x) → −∞ as x → 0. The other important properties of the logarithm function are: ln(xy) = ln x + ln y

for

1

y = ln(x)

0 1

2

3

4

-1

x, y > 0 Figure 3.3: Graph of ln(x).

1 ln( ) = − ln x x

for

x>0

These can be proved as follows. Let x and y be two positive values. We know that there are values a and b such that x = exp a and y = exp b. So ln(xy) = ln(exp a exp b) = ln(exp(a + b)) = a + b = ln x + ln y Secondly, exp(−a) = 1/ exp(a) = 1/x. So 1 ln( ) = ln(exp(−a)) = −a = − ln x x We can get the graph of ln x simply by turning round the graph of ex . It is given in Fig 3.3.

56

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

Derivative of ln x

3.3.2

We do this in the same way as we did the inverse trig functions. Let y = ln x. Then, by definition, x = ey . Differentiate this wrt x, using the Chain Rule: 1 = ey So

dy dx

1 1 dy = y = dx e x

Therefore 1 d ln(x) = dx x This is important.

3.4

Hyperbolic Functions

There are some simple combinations of exponential functions that have been given special names. There is really nothing much to them at this level, but they should be known. The functions sinh x, cosh x and tanh x are defined as follows cosh x = 12 (ex + e−x ) sinh x = 12 (ex − e−x ) sinh x tanh x = cosh x They are pronounced as ‘cosh’, ‘shine’ and ‘tansh’ (or ‘than’) and are called the Hyperbolic Trigonometric Functions. You may have buttons for them on your calculator. Their properties can be deduced easily from those of ex . You should check that cosh(−x) = cosh(x)

sinh(−x) = − sinh(x) cosh(0) = 1

as

x → ±∞

d cosh x = sinh x dx

tanh(−x) = − tanh(x)

sinh(0) = 0    cosh x → ∞ sinh x → ±∞   tanh x → ±1 d sinh x = cosh x dx

cosh2 x − sinh2 x = 1 There are also formulas corresponding to all the usual trigonometric formulas, e.g. for sinh(x + y) and sinh 2x. They are the same as the trig formulas except for some of the signs.

3.4. HYPERBOLIC FUNCTIONS

3.4.1

57

More Differentiation Examples

Now we can do derivatives with logs in, and with hyperbolic functions. Example 3.6. f (x) = x ln x By the product rule f 0 (x) = 1. ln x + x.

1 = ln x + 1 x

Example 3.7. f (x) = ln2 x This is chain rule. Let y = u2 where u = ln x. Then dy du 1 2 dy = = 2u. = ln x dx du dx x x Example 3.8. f (x) = ln(1 + x2 ) Chain Rule. Let y = ln u where u = 1 + x2 . Then dy du 1 2x dy = = .2x = dx du dx u 1 + x2 Example 3.9. f (x) = sinh3 (1 + x2 ) Chain Rule. Let y = u3 where u = sinh v, where v = 1 + x2 . Then dy du dv dy = = 3u2 . cosh v.2x = 6x sinh2 (1 + x2 ) cosh(1 + x2 ) dx du dv dx

Questions 10

(Hints and solutions start on page 71.)

Q 3.1. To get used to using the right buttons on your calculator, check the following values: e2.3 ≈ 9.974,

e−1 ≈ 0.3679,

ln(2) ≈ 0.6931,

ln(1000000000) ≈ 20.7233

(If you got ln 2 = 0.3010 then you are using the wrong logarithm button. You are using ‘common logarithms’ ln10 2.). What is the value of ln 101000 ? Q 3.2. Differentiate the following functions g(x) = 2 ln x − ex ,

f (x) = 2ex + x, k(x) =

ln x , x2 2x

n(x) = 2e r(x) =

ex − e−x , ex + e−x

1 l(t) = p(− t2 ), 2 x2

−e ,

h(x) =

ex − 1 ex + 1

m(s) = ln(1 + s2 ) 2

x

p(x) = ln(e ),

s(χ) = ln(

χ2 + 2χ + 1 ), χ2 − 2χ + 1

ex q(x) = x e

t(x) = ln(ln x)

58

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

Q 3.3. Differentiate the following functions f (x) = cosh(2x),

g(x) = 2 cosh x − 3 sinh x,

l(x) = sinh(sin x),

h(x) =

1 , cosh x

k(x) =

sinh x cosh x

n(x) = sinh2 (x2 )

m(x) = cosh(ln x),

Q 3.4. Work out the equation of the tangent at the point (a, ea ) on the graph y = ex . What condition does a have to satisfy if this tangent is to pass through the origin (0, 0)? Draw a picture to illustrate your answer. Now find the values of m for which the straight line y = mx fails to meet the graph y = ex . Q 3.5. Show that x(t) = ept , where p is a constant, only satisfies the equation a¨ x +bx+cx ˙ = 2 0 if ap + bp + c = 0. Q 3.6. Show that x(t) = e2t (sin(t) − cos(t)) satisfies the equation x¨ − 4x˙ + 5x = 0 for all values of t. Q 3.7. For what values of the constant µ does the function x(t) = et cos(µt) satisfy the equation x¨ − 2x˙ + 10x = 0 for all values of t? Q 3.8. Use the properties of the logarithm, in particular ln ab = ln a + ln b, ln a/b = ln a − ln b and ln xa = a ln x, to simplify the following expressions as much as you can — you can then differentiate them. y = ln((1 + x)2 (1 + 2x)3 ),

y = ln((1 + x)(2 + x)),  y = ln

1−x 1+x

r

2 ,

y = ln

y=

ln((1 + x)2 ) ln(1 + x)

1−x 1+x

Q 3.9. This question functions. Show that √ √ is about the inverse functions of the hyperbolic cosh−1 (x) = ln(x + x2 − 1) if x ≥ 1 and sinh−1 (x) = ln(x + x2 + 1) (which is a genuine inverse). *Q 3.10. This goes back to a point raised in the text. I claimed, at one point, that if we have constants A and B such that A sin x+B cos x = 0 for every value of x then A = B = 0. Prove this (remembering that ‘every’ automatically includes ‘any’). Now prove the same kind of result for A sin x + B cos x + C sin 2x + D cos 2x = 0 and for A + Bex + Ce2x = 0.

3.5. POWERS

59

*Q 3.11. Consider the equation dy = ayandy(0) = 1 dx

(1)

where a 6= 0 is a constant. Define a new variable z by z = ax and show that dy = yandy(0) = 1 dz Show that the single solution to (1) is y = eax . Did I really need to put on the restriction a 6= 0? *Q 3.12. This is an alternative view of the exponential function. Suppose I invest a sum P for one year at an interest rate of I%. Explain why, at the end of the year, I will have sum (1 + k)P where k = I/100. Now suppose that instead of getting I% per annum I get I/2% every six months. What sum do I now have at the end of one year? Now suppose that I get I/N% interest added N times during the year. Give a formula for the total sum at the end of one year. Now use your calculator to work out some numbers. Take a capital P = 1 an interest rate of 10% and work out the sum at the end of one year for N = 2, 10, 100, 1000, 10000. Do you notice anything? This all boils down to the following theorem about the exponential function (which I am not asking you to prove!).  x N lim 1 + = ex N →∞ N

3.5

Powers

Let n be a positive integer. Then xn = x.x.x . . . x n times. So ln(xn ) = ln(x.x.x . . . x) = ln(x) + ln(x) + · · · + ln(x) = n ln(x) (for example, ln(x3 ) = ln(x.x.x) = ln(x) + ln(x) + ln(x) = 3 ln(x).) Now consider x1/n . (x1/n )n = x so

ln((x1/n )n ) = ln x so

Therefore ln(x1/n ) =

1 ln x n

Finally, by combining the two results, we get ln(xn/m ) =

n ln x m

n ln(x1/n ) = ln x

60

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

Consequently, xn/m = exp(

n ln x) m

because

n ln x) = exp(ln(xn/m )) = xn/m m This looks ridiculously complicated, but is extremely useful. Firstly, it the the way that your calculators evaluate non-integral powers. If you ask your calculator to evaluate 3.451.23 what it actually evaluates is exp(

exp(1.23 ln 3.45)

2

√ Secondly, 2 π

this gives us a simple way to say what we actually mean by expressions like or e . We are just going to say that for any x and a, with x > 0, xa = ea ln x

This is our definition of a power. We have seen that it is consistent with our normal ideas of powers. As an exercise you can now prove all the usual ‘rules of indices’ by using this definition. Theorem 3.13. For x > 0 and for any value of a d a x = axa−1 dx Proof. We claimed this earlier on, now having a definition we can prove it. It is a simple use of the chain rule. xa = exp(a ln x) so

d a d x = exp(a ln x) dx dx

Now we go through the chain rule motions: y = exp u where

u = av

where

v = ln x

dy dy du dv 1 xa = = exp(u).a. = a = axa−1 dx du dv dx x x

3.6

* Irrational Numbers

I commented in the last section that there is no great difficulty in understanding what is meant by a power and logarithm functions to make easy sense of a power like like x3/7 but that we needed the exponential √ √ 2 Why? Can we not just write 2 as a fraction and use the old definition? No we can’t. It turns out x .√ that 2 is one of a class of numbers that cannot be written as fractions (ratio of two whole numbers). They are called the Irrational Numbers and the ones that can be written as fractions are called the Rational Numbers.

3.6. * IRRATIONAL NUMBERS

61

The following numbers are certainly Rational Numbers. −13 , 23

7 , 2

56,

123.456 (=

123456 ) 1000

It is not√always easy to decide whether a number is Irrational. Let me show you the argument that proves that 2 must be Irrational. This proof is in Euclid’s Elements (published circa 300 B.C.) and is beautifully simple. √ Theorem √ 3.14. 2 is an Irrational Number. Or, equivalently, we cannot find whole numbers p and q such that 2 = p/q. Proof. Once more we use ‘proof by contradiction’ (we used it when developing the properties √ of the exponential function). We assume that it is possible to find whole numbers p and q such that 2 = p/q and then try to get ourselves into a logical tangle. We are only interested in the ratio of p and q so any common factors that the numbers have can be cancelled out. Suppose that this has been done. Then p and q cannot both be even numbers (or there is a common factor of 2). Squaring up and rearranging, we get 2q 2 = p2 . It is at once clear that p2 must be an even number (it’s twice something). In that case p must be an even number as well because the square of an odd number is still odd. So we can choose to write p = 2r, where r is another whole number. Put this back into the earlier equation and get 2q 2 = 4r2 or q 2 = 2r2 . So, as before, we see that q must be an even number. So both p and q are even numbers. But we started with the assumption that they were not both even! √ This gives us a contradiction and so we can conclude that 2 cannot be written as a fraction and hence is an Irrational Number. It is not recorded who it was first realised that not all numbers are fractions. We do know that, whoever he was, he probably lived in the fourth century BC. There is an ancient tradition that says that he was murdered for his pains. Why murder somebody for making a mathematical discovery? The answer is that he was probably a member of (or associated with) the religious cult started by Pythagoras (of the triangle). The Pythagoreans seem to have had a theory of the world that was based on whole numbers and consequently on their ratios. The discovery that whole number ratios were not everything seems to have put the cat amongst the pigeons. Eliminating the discoverer was a predictable reaction, though not nowadays regarded as an acceptable form of mathematical proof. Here are some Irrational Numbers √ 2,

√

3,

21/3 ,

π,

e,

ln 2

√ (in fact the square root of a whole number is either another whole√number, like 4, or else irrational). As an exercise, try adapting √ my earlier proof to prove that 3 is irrational. Then check that your proof does not also prove that 4 is irrational, because it ain’t. The difference between rational and irrational numbers shows up nicely in their decimal expansions. The decimal expansion of a rational number is always eventually periodic (repeats itself). 1 3 2 7 1 6 1 2

= 0.333333333333333333 . . . = 0.285714285714285714 . . . = 0.166666666666666666 . . . = 0.500000000000000000 . . .

62

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

On the other √ hand the decimal expansion of an Irrational number can never become periodic.The decimal expansion of 2 starts off as 1.4142135623730950488016887242096980785696718753769480731766797 379907324784621070388503875343276415727350138462309122970249248 360558507372126441214970999358314132226659275055927557999505011 527820605714701095599716059702745345968620147285174186408891986 095523292304843087143214508397626036279952514079896872533965463 —not that that proves anything, the period might have been big.

Questions 11

(Hints and solutions start on page 72.)

*Q 3.15. As I said, prove that

√ 3 is an irrational number. Now prove that the cube root of 2 is irrational.

*Q 3.16. How would you use geometric series to work out the value of the following periodic ‘decimal’ as an exact fraction? 0.123123123123123 . . . What’s the answer? What is the value of 0.11112121212121 . . .? Now go on and give a reasonably coherent argument to prove that any periodic (or eventually periodic) decimal must represent a rational number? *Q 3.17. X is any point on the x-axis and  is any positive quantity (as small as you like). Prove that there must be a rational number within  of X. Prove that there must be an irrational number within  of X as well. Deduce that between any two different numbers there are infinitely many rational numbers and infinitely many irrational numbers.

Appendix A Some Useful Tables Table of Derivatives Here, for reference, is a table of the basic derivatives that you have met so far. f (x)

f 0 (x)

Powers x

1

xa

axa−1

Trigonometric Functions sin(x)

cos(x)

cos(x)

− sin(x)

tan(x)

sec2 (x)

cosec(x)

− cos(x)/ sin2 (x)

sec(x)

sin(x)/ cos2 (x)

cot(x)

− cosec2 (x)

a any constant (x in radians)

Inverse Trig Functions √ arcsin(x) 1/ 1 − x2 √ arccos(x) −1/ 1 − x2 arctan(x)

1/(1 + x2 )

Exp & Log eax

aeax

a any constant

ln(x)

1/x

x>0

sinh(x)

cosh(x)

cosh(x)

sinh(x) 63

64

APPENDIX A. SOME USEFUL TABLES

Greek Letters Mathematicians have a habit of using Greek letters in their work. This is partly to show that they are educated and partly because there are not enough Roman letters to go round (Mathematicians don’t really like the Computing habit of using words for quantities — because of the problem with multiplication: is CAT a thing with a tail or C.A.T?). This can be a nuisance if you are not familiar with the Greek alphabet. So here it is as a reference, together with the pronunciations of the names of the letters. Capital Name Pronunciation of name α

A

alpha

alfa

β

B

beta

beeta

γ

Γ

gamma

gamma like gammy

δ

∆

delta

as in river



E

epsilon

epsuylon or epsi-lon

ζ

Z

zeta

zeeta

η

H

eta

eat a (cake)

θ

Θ

theta

th as in thing not the

ι

I

iota

I owe Ta

κ

K

kappa

cap a(long)

λ

Λ

lambda

lamda

µ

M

mu

mew like cat

ν

N

nu

new like old

ξ

Ξ

xi

zi

o

O

omicron

short ‘o’

π

Π

pi

pie

ρ

P

rho

rho(dodendron)

σ

Σ

sigma

sigma

τ

T

tau

ta(lk)

υ

Υ

upsilon

oopsilon

φ

Φ

phi

fi(ne) (UK) fee (US)

χ

χ

chi

ki(nd)

ψ

Ψ

psi

(shar)p si(ght)

ω

Ω

omega

alpha and omega

Appendix B Solutions to Exercises Solutions for Questions 1

(page 8).

Solution 1.1: The Newton Quotient is ((x + h)3 − x3 )/h. Now (x + h)3 = x3 + 3hx2 + 3h2 x + h3 , as you can check by multiplying out. So, subtracting x3 and dividing by h we get 3x2 + 3hx + h2 . The limiting value of this as h → 0 is 3x2 , so this is the required derivative. Solution 1.2: The Newton Quotient is (1/(1 + x+ h) −1/(1 + x))/h = −1/(x+ 1)(x+ 1 +h). As h → 0 we have (x + 1 + h) → (x + 1). So the derivative is −1/(x + 1)2 . Solution 1.3: For example, with h = 0.05, (sin(0.05) − 0)/0.05 = 0.99958. Solution 1.4: Consider the function |x| at x = 0, draw a picture.

Solutions for Questions 2

(page 13).

Solution 1.5: 4x3 , 5x4 + 12 , 24t5 − 15t4 + 2, 4p3 − 9p−4 − 2p−2 . Solution 1.6: p0 (x) = 2 cos x, q 0 (x) = sec2 x + cos x, r 0 (θ) = −3 sin θ − 4 cos θ, m0 (λ) = ˙ 2λ − 4 cos λ + 3 sin λ, f 0 (x) = 12 x−1/2 + 12 x−3/2 , h(t) = 15 t1/4 − 2t−2/3 − 2t−3 , φ0 (x) = 4 1 −1/2 2 . sec x − 3 sin x − 2 x Solution 1.7: f 0 (x) = sin x + x cos x, g 0 (x) = − sin2 x + cos2 x, r 0 (x) = −2 cos x sin x, p0 (x) = 1 + 2 sin x + 2x cos x, q 0 (x) = 4 sin x cos x, u0 (x) = v 0 (x) = 2 sin(x)/ cos3 (x), t9/4 − 17 t13/4 . w(t) ˙ = 2t − 3t2 + 13 4 4 Solution 1.8: p0 (x) = −

u0 (x) =

1 , x2

−4x , 2 (x − 1)2

r 0 (x) =

1 − x2 , (1 + x2 )2

e(t) ˙ =

s0 (x) =

2 − 2t2 , (t2 − t + 1)2 65

7 , (2x + 3)2

m(t) ˙ =√

t0 (x) =

−1 √ , t(1 + t)2

3 − 2x − 2x2 . (x2 − x + 1)2 l0 (x) =

1 . (x + 1)2

66

APPENDIX B. SOLUTIONS TO EXERCISES

v 0 (x) =

1 , (1 + cos x)

w 0 (x) =

1 , (1 − sin x)

z 0 (x) =

− cos x , sin2 x

f 0 (x) = sec2 x − cosec2 x.

Solution 1.9: y 0 = 3x2 . So the slope at (1, 1) is 3. So y − 1 = 3(x − 1) or y = 3x − 2. This meets the x-axis when y = 0, so x = 2/3. Solution 1.10: y 0 = cos(x). You know that cos(π) = −1 and sin(π) = 0. So tangent is y − 0 = −1.(x − π) or x + y = π. This meets the axes at (π, 0) and (0, π). Solution 1.11: Derivative is −1/x2 . So the slope at P is −1/a2 . The equation of the tangent is therefore y − 1/a = −1/a2 (x − a). This meets the x-axis at x = 2a and the y-axis at y = 2/a. These give the base and height of a right-angled triangle. The area is therefore A = 12 (2a)(2/a) = 2. Note that the area does not depend on the value of a, i.e. on the choice of the point P. Solution 1.12: The product of the slope of the tangent and the slope of the normal must be −1. In this case the slope of the tangent is 2a and so the slope of the normal must be −1/(2a). So the equation of the normal is y − a2 = −1/(2a)(x − a). It meets the y-axis at y = a2 + 1/2. Solution 1.13: f (x) = 2x − 3x2 + x3 so f 0 (x) = 2 − 6x + 3x2 . We just want to know when this is positive and when it is negative. It is a quadratic with roots 1 ± 3−1/2 . Between these the derivative is negative, so the value of the function is decreasing. Outside, the derivative is positive so the function is increasing. Solution 1.14: The function fails to be defined only at x = ±1 (where the denominator becomes zero). The derivative works out as −4x/(x2 − 1)2 and this has the opposite sign to x. So the graph is always increasing when x < 0 (and not = −1) and is always decreasing when x > 0 (and not = 1). Solution 1.15: The slope at (a, a2 ) is 2a, so the slope at Q is −1/(2a) and hence Q is the point (b, b2 ) where b = −1/(4a). Now work out the equations of the tangents and solve them simultaneously. You should get y = −1/4 regardless of the value of a. √ √ Solution 1.16: Multiply ‘top & bottom’ by a + b to get the formula. For the derivative: √ 1 1 (x + h) − x 1 √ ( x + h − x) = √ √ =√ √ . h h x+h+ x x+h+ x √ The final denominator tends to 2 x as h → 0.

Solutions for Questions 3

(page 20).

Solution 1.18: f (g(x)) = 1 + (1/x2 ), g(f (x)) = 1/(1 + x2 ), g(g(x)) = x. Solution 1.19: f (g(h(x))) = 1 + 1/(1 − x), f (h(g(x))) = 2 − 1/x, g(h(f (x))) = −1/x, etc. Solution 1.20: 10(2x + 1)4 ,

(3x − 2)−2/3 ,

3 sin2 x cos x,

−10 cos9 x sin x,

√ x/ 1 + x2 ,

67 5 cos 5x,

−3 sin(3x + 1),

2 3 (x + 2x − 6)−1/3 (3x2 + 2), 3 6 sin2 (x) cos x + 9 cos2 x sin x,

2x sec2 (1 + x2 ),

− sin(x) cos(cos x)), √ 1 sec2 (x)/ tan x, 2

−(1 − 1/t2 ) sin(t + 1/t),

√ 1 (cos x + sin x)/ sin x − cos x, 2

√ √ 1 cos( sin t) cos t/ sin t, 2

√ √ t−1/2 sin( t) cos( t),

2 x cos(x2 ) sin−2/3 (x2 ). 3

Solution 1.21: f (f (x)) = (x2 + c)2 + c = x4 + 2cx2 + c2 + c, and so on Solution 1.22: r

1 1−x · , 1 + x (1 − x)2

−2x cos(x2 )/ sin2 (x2 ),

−4 cos(x) sin3 (x)/(1 + sin4 x)2 ,

1 − 4x/(x2 − 1)2 ,

1/(1 + x2 )3/2 , 0, and so on . . . .

Solution 1.23: Differentiating the LHS gives 1+2x+3x2 +4x3 +· · ·+nxn−1 . Differentiating the RHS gives nxn+1 − (n + 1)xn + 1 . (x − 1)2 Putting x = −1 gives 1 − 2 + 3 − 4 + · · · ± n on the LHS and the RHS becomes 1 − (2n + 1)(−1)n n(−1)n+1 − (n + 1)(−1)n + 1 = . 4 4 Solution 1.24: c(s(s(s(x)))).c(s(s(x))).c(s(x)).c(x), where c, s stand for cos and sin. Solution 1.25: p(x) = ln ln sin(x). sin x takes values between −1 and 1. ln sin(x) is only defined if sin(x) > 0 and then takes values ≤ 0. But that means that the final logarithm is not defined at all! So p(x) is not defined for any value of x. If we ignore this and differentiate like a machine we get the derivative cot(x)/ ln(sin(x)). This is certainly defined for quite a lot of values of x! Solution 1.26: No solution given. Solution 1.27: No solution given.

68

APPENDIX B. SOLUTIONS TO EXERCISES

Solutions for Questions 4

(page 25).

Solution 1.28: The derivatives, starting from f (x) are sin(x),

cos(x),

− sin(x),

− cos(x),

sin(x)

and now they start to repeat. In fact the sequence of derivatives goes through a cycle of four functions. To find the 300th derivative we just have to decide where 300 comes in the cycle. Now 300 is a multiple of 4 and therefore the required derivative is sin(x). The next problem is harder. Let me put down a number of derivatives x sin x,

sin x + x cos x,

−3 sin x − x cos x,

2 cos x − x sin x

−4 cos x + x sin x,

5 sin x + x cos x

Now you have to try to spot the pattern. There is still a kind of cycle of 4, except that the numerical coefficient of the first term is steadily increasing. The 300th derivative is −300 cos x + x sin x. x−7/2 , 105 x−9/2 . It should by now be easy to Solution 1.29: 12 x−1/2 , − 14 x−3/2 , 38 x−5/2 , − 15 16 32 see what is happening. The problem is to turn it into a formula. The power of x in the nth derivative is −(2n − 1)/2. If n is even the sign is negative and if n is odd the sign is positive. The denominator of the coefficient is 2n and the numerator is 1.3.5.7.9...(2n − 3). This actually works out to give f (n) (x) = (−1)n+1

2(2n − 2)! −(2n−1)/2 x 4n (n − 1)!

— so there! Solution 1.30: We just have to find out the second derivative of x(t). x(t) ˙ = Aω cos(ωt + α),

x¨(t) = −Aω 2 sin(ωt + α) = −ω 2 x(t)

Solution 1.31: This is just hard work. Work out the values of y 0 and y 00 , plug them into the equation together with y and hope that the resulting mess simplifies down to 0. It probably easiest if you write y as a product, rather than a quotient. Here are clues to help you check. cos 2x + sin 2x = x−1 cos 2x + x−1 sin 2x x y 0 = −x−2 cos 2x − 2x−1 sin 2x − x−2 sin 2x + 2x−1 cos 2x y 00 = 2x−3 cos 2x + 2x−3 sin 2x4x−2 sin 2x − 4x−2 cos 2x − 4x−1 cos 2x − 4x−1 sin 2x. y=

Now compute y 00 + 2x−1 y 0 + 4y. Solution 1.32: It must have been a polynomial.

69 Solution 1.33: (f g)00 = f 00 g + 2f 0 g 0 + f g 00 , (f g)000 = f 000 g + 3f 00 g 0 + 3f 0 g 00 + g 000 and so on. The layout is the same as for (x + y)n . Solution 1.34: No solution given. √ Solution 1.35: y 0 = ny/ 1 + x2 , y 00 = n2 y/(1 + x2 ) − nxy/(1 + x2 )3/2 Solution 1.36:b fn (x) differentiates to give nfn−1 (x) if n > 1. So fn can be differentiated n − 1 times.

Solutions for Questions 5

(page 26).

Solution 1.37: 1/x → 0 as x → ∞, so cos(1/x) → 1. Solution 1.38: Use common sense on the first one. The numerator is constant and the denominator is getting bigger and bigger. In fact the denominator is tending to infinity. So the fraction tends to 0. In the second limit both the numerator and the denominator are tending to infinity, so the result is not quite so obvious. If you plug in a few values you will soon convince yourself that the limiting value is 1. The easy way to see this is to write the function as (1 + 1/x2 )/(1 − 1/x2 ). In this form both numerator and denominator √ √tend x + 1 − x= to 1√in the limit, so we were right. For the last one, following the hint, √ 1/( x + 1 + x) which makes it obvious that the limit is 0. Solution 1.39: f (x) = x + 1 for any value of x other than 1. So the limiting value is clearly 2.

Solutions for Questions 6

(page 32).

Solution 2.1: f −1 (x) = (x − 4)/2, g −1 (x) = (2 − x)/5. Solution 2.2: The inverse is also f (x) = 1/x. Solution 2.3: The inverse g(x) is given by: if x ≥ 1 then g(x) = x − 1 and if x < 1 then g(x) = 2(x − 1). Solution 2.4: The graph must be symmetrical about the line y = x — i.e. if (x, y) is on the graph then so is (y, x). Solution 2.5: f is not defined at x = −d/c. It cannot take the value a/c. Suppose we have f (x) = f (y) (with neither x nor y making the denominator zero). Then (ax + b)(cy + d) = (ay + b)(cx + d) or (ad − bc)(x − y) = 0. So since ad 6= bc we must have x = y. The inverse is f −1 (x) = (b − dx)/(cx − a).

Solutions for Questions 7

(page 37).

Solution 2.9: No solution given. Solution 2.10: π/2, −π/2, 0, π, π/2, 0, π/4, π/4, π/3.

70

APPENDIX B. SOLUTIONS TO EXERCISES

√ √ Solution 2.11: (1, 0), (1, π/2), (1, π), (1, −π/2), ( 2, π/4), ( 2, 3π/4) and √ ( 2, −3π/4). p √ √ , arctan(x) + x/(1 + x2 ), 2/ 1 − 4x2 , −3/ 1 − (3x + 1)2 , Solution 2.12: 5/ 1 − x2√ 2x/(2 √ + 2x2 + x4 ), ±1/ 1 − x2 (+ if x < 0 and − if x > 0, not diff at x = 0), −1/(2 x − x2 ). Solution 2.13: You should get 0 because f (x) = π/2. Solution 2.14: You should get a ‘sawtooth’ graph in both cases. √ Solution 2.15: y 0 = (a + 2 arcsin(x))/ 1 − x2 , y 00 = 2/(1 − x2 ) + (ax + 2x arcsin(x))/(1 − x2 )3/2 . Now plug in. Solution 2.16: No solution given.

Solutions for Questions 8

(page 41).

Solution 2.17: (1) 2x + 1 + 2yy 0 = 0, Put x = 0, y = 1 and get y 0 = −1/2. So equation of tangent is y − 1 = −x/2. (2)3x2 + 3y 2y 0 = 0. Put x = 2, y = 1 and get y 0 = −4. So equation of tangent is y − 1 = −4(x − 2). (3) 3x2 + y + xy 0 + 3y 2y 0 = 0. Put x = 1 and y = 0 and get y 0 = −3. Tangent is y = −3(x − 1). (4) (1 + y 0) cos(x + y) − (1 − y 0) sin(x − y) = 0. x + y = 0, x − y = π/2. So y 0 = 0 and tangent is y = −π/4. (5) After simplifying we get x4 + y 4 = 2. Differentiate: x3 + y 3y 0 = 0. So y 0 = −1 and tangent is y − 1 = 1 − x or y = 2 − x. Solution 2.18: No solution given. Solution 2.19: No solution given.

Solutions for Questions 9

(page 44).

Solution 2.20: (1) x˙ = 2t, y˙ = 3t2 , so dy/dx = 3t/2 = 3/2. (2) x˙ = − sin t, y˙ = cos t, so dy/dx = − cot t = −1 (3) x˙ = 2λ, y˙ = 2λ, so dy/dx = 1. (4) x(1/2) ˙ = −32/25, y(1/2) ˙ = 24/25 so dy/dx = −3/4. Solution 2.21: x˙ = 2t, y˙ = 3t2 , dy/dx = 3t/2. x(−t) = x(t) ≥ 0, y(−t) = −y(t). Curve is symmetrical about x-axis and has slope 0 at origin, where it has a cusp. Solution 2.22: The only point on the circle that you do not get is (−1, 0), which is what corresponds to t = ±∞. Solution 2.23: No solution given. Solution 2.24: No solution given. Solution 2.25: No solution given.

71

Solutions for Questions 10

(page 57).

Solution 3.1: No solution given. Solution 3.2: f 0 (x) = 2ex + 1, g 0 (x) = 2/x − ex , h0 (x) = 2ex /(ex + 1)2 , k 0 (x) = (1 − 2 2 2 ln x)/x3 , l0 (x) = −te−t /2 , m0 (x) = 2s/(1 + s2 ) n0 (x) = 4e2x − 2xex , p0 (x) = 1, 2 q 0 (x) = (2x − 1)ex −x , r 0 (x) = 4e2x /(1 + e2x )2 , s0 (x) = −4/(χ2 − 1), t0 (x) = 1/(x ln x). Solution 3.3: f 0 (x) = 2 sinh(2x),

g 0 (x) = 2 sinh x − 3 cosh x,

k 0 (x) = 1/ cosh2 x,

l0 (x) = cos(x) cosh(sin x),

h0 (x) = − sinh(x)/ cosh2 (x) m0 (x) = (1 − (1/x2 ))/2

n0 (x) = 4x sinh(x2 ) cosh(x2 ) Solution 3.4: Slope is ea . Tangent is y − ea = ea (x − a). This is satisfied by (0, 0) if −ea = −aea or a = 1. Point is (1, e) and slope is e. Your picture should make it clear that y = mx fails to meet the graph if 0 ≤ m < e. x + bx˙ + cx = (ap2 + bp + c)ept Since the Solution 3.5: x˙ = pept and x¨ = p2 ept . So a¨ exponential function is never zero this can only be zero if ap2 + bp + c = 0. Solution 3.6: Hard pounding! x˙ = 2e2t (sin t − cos t) + e2t (cos t + sin t) = e2t (3 sin t − cos t) x¨ = 2e2t (3 sin t − cos t) + e2t (3 cos t + sin t) = e2t (7 sin t + cos t) So, putting it all together, x¨ − 4x˙ + 5x = e2t ((7 − 12 + 5) sin t + (1 + 4 − 5) cos t) = 0 So the equation is satisfied for all t. Solution 3.7: Same process x˙ = et cos µt − µet sin µt = et (cos µt − µ sin µt) x¨ = et (cos µt − µ sin µt) + et (−µ sin µt − µ2 cos µt) = et ((1 − µ2 ) cos µt − 2µ sin µt) Putting it all together x¨ − 2x˙ + 10x = et ((1 − µ2 − 2 + 10) cos µt + (−2µ + 2µ) sin µt) = et (9 − µ2 ) cos µt This will certainly be identically zero if µ = ±3. et is never zero and cos µt can only be zero at certain points. So the equation can only be satisfied for all values of t if µ = ±3. Solution 3.8: (1) y = ln(1 + x) + ln(2 + x), so y 0 = 1/(1 + x) + 1/(2 + x). (2) y = 2 ln(1 + x) + 3 ln(1 + 2x), so y 0 = 2/(1 + x) + 6/(1 + 2x). (3) y = 2 ln(1 + x)/ ln(1 + x) = 2,

72

APPENDIX B. SOLUTIONS TO EXERCISES

so y 0 = 0. (4) y = 2 ln(1 − x) − 2 ln(1 + x), so y 0 = −2/(1 − x) − 2/(1 + x). (5) y = 12 (ln(1 − x) − ln(1 + x)), so y 0 = 12 (−1/(1 − x) − 1(1 + x)). Solution 3.9: We follow the usual procedure, but have to be careful about signs. I will just do cosh−1 . The other goes the same way. If y = cosh−1 (x) and x ≥ 1 then x = cosh(y) = by ey to get e2y − 2xey + 1 = 0 which is a quadratic for ey . (ey + e−y )/2. Multiply through √ Its solutions are ey = x ± x2 − 1. We choose to have y ≥ 0, so ey ≥ 1, so we must pick the + sign. Solution 3.10: Put x = 0 and get B = 0; put x = π/2 and get A = 0. The next one can be done by looking at x = 0, π/2, π. For the last one, it helps to differentiate once or twice before you start putting in values. Solution 3.11: dy/dx = dy/dz.dz/dx = a.dy/dz. So the original equation becomes dy/dz = y with condition y(0) = 1. We know that the solution to this is y = ez , so the equation in its original form has solution y = eax . Solution 3.12: No solution given.

Solutions for Questions 11

(page 62).

√ Solution 3.15: Suppose that 3 = p/q as a fraction in its lowest terms. Then p2 = 3q 2 . This means that 3 divides p2 . Convince yourself that this means that 3 must divide p as well, say p = 3k. Then√9k 2 = 3q 2 so q 2 = 3k 2 and therefore 3 divides q. Contradiction. (This does not work for 4 because knowing that 4 divides p2 does not imply that 4 divides p (e.g. p = 6). The cube root argument goes along in much the same way as the square root one. Solution 3.16: The number is 1/10 + 2/102 + 3/103 + 1/104 + 2/105 etc. Because of the periodicity we can arrange this as (1/10 + 2/102 + 3/103) + (1/10 + 2/102 + 3/103 )/103 + (1/10 + 2/102 + 3/103 )/106 + · · · So the number is

a(1 + b + b2 + b3 + b4 + · · · ) where a = 0.123 and b = 1/103. The Geometric series sums to give 1/(1 − b) = 1000/999. So the number is 123/999. The number 0.121212 . . . works out to give 4/33, so the number in the question is 3667/33000. The general argument just follows the ideas of the examples. Gather a single periodic block together and get it as the coefficient of a geometric series in powers of 10. Then add on the header bit, which certainly gives a fraction. Solution 3.17: The fractions 1/2n can be made as small as we like by making the whole number n big enough. Choose n so that 1/2n is less than the gap between x −  and x + . Then one of the numbers 1/2n , 2/2n , 3/2n , . . . √must land in the gap, and these are all rational numbers. For the second bit, note that k 2/2n is irrational if k 6= 0 and then argue as before.