p1 chapter 12 & 15 cie centre a-level pure maths © adam gibson

P1 Chapter P1 Chapter 12 & 1512 & 15

CIE Centre A-levelCIE Centre A-level Pure Maths Pure Maths

© Adam Gibson

Extending differentiation

In Chapter 11 we studied composite functions

( ( ))f g x

A simple example:8

( ) 3 4, domain

( ) , domain

g x x x

f x x x

What is df

dx? What is dg

dx?

How about ( ( ))df g x

dx?

BORING!

Or a “function of a function”

The chain rule - Introduction

g f0x 02x 06x

Let’s add 1 to x so we can find the gradient

g f0 1x 02 2x 06 6x

Of course this example is easy; but it shows that: the rate of change of f(g(x)) with respect to x is equal to the rate of change of g with respect to x multiplied by the rate of change of f with respect to g

gradient=2 gradient=3

The chain rule - continued

Mathematically:.

df df dg

dx dg dx

This looks obvious, but remember

You cannot do this: .df df dg

dx dg dx

See p. 183 and p. 184 for a careful discussion of thederivation of the chain rule

“The Chain Rule”

The chain rule – Examples

Let’s go back to our first example8

( ) 3 4, domain

( ) , domain

g x x x

f x x x

3dg

dx 78

dfg

dg

7 78 .3 24(3 4)df df dg

g xdx dg dx

Use the chain rule:

Much easier than using the binomial theorem!!

8( ( )) (3 4)f g x x

The chain rule – Examples

This rule allows to differentiate more complicatedfunctions. Here’s an example:

3 2

1( )

( 2 5)f x

x x

What is … ?

df

dx

Express f(x) as a composite function.

3 2

1( )

( ) 2 5

f hh

h x x x

( ) ( ( ))f x f h x

2

1df

dh hWe know: and:23 4

dhx x

dx

The chain rule – Examples

Now we can apply the chain rule:

22

2

3 2 2

.

1 .(3 4 )

3 4

( 2 5)

df df dh

dx dh dx

x xh

x x

x x

Q: What are the stationary points of f ?

The chain rule is very important in real world applicationsof calculus, because we are often dealing with severalquantities (e.g. weather - time, temperature, pressure, density), many of which vary depending on each other.

A real world example

Remember our coffee cup?Question : how fast is the level of the coffee rising?

-6

-4

-2

0

2

4

6

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

2

( ) 3, 02

hr h h

h

r

Coffee

A real world example - continued

It can be shown (using integration) that the volume abovethe half way point varies with the height as follows:

5 3( ) 2 1810

V h h h h

The cup is already half full. If we also know that the coffee is being poured at a rate of 42 cm3 per second, then at what rate will the height be increasing when h = 1cm?

The question asks us to find dh

dt

The Chain Rule tells us: dV dV dh

dt dh dt

Second derivatives (Chapter 15)

0

2

4

6

8

10

12

14

16

-6 -5 -4 -3 -2 -1 0 1 2 3

Let’s start by looking at our old friend, the quadratic curve:

2 3 5x x

Sketch the curve on paper, and sketchthe derivative on thesame axes

Second derivatives continued

What is the equation of the derivative? 2 3y x Now draw the derivative of the derivative

-10

-5

0

5

10

15

20

-6 -5 -4 -3 -2 -1 0 1 2 3

quadratic

derivative

2nd deriv2y

Second derivatives – definitions and notation

The derivative of the derivative is usually calledthe second derivative

It’s written like this: 2

2

d y

dxor sometimes ''( )y x

2

2

d y

dx

Don’t get confused about this.We don’t write:

2

2

dy

dxor

2

2

d y

d x

What we are really doing is applying the operator d

dxtwice; it’s similar to a composite function.

Second derivatives continued

Sketch the first andsecond derivatives of:

3 23 3 8y x x x

-30

-20

-10

0

10

20

30

-3 -2 -1 0 1 2 3

cubic

derivative

second deriv

The first derivative of a quadratic is a :

So far we have learned …So far we have learned …

The second derivative of a quadratic is a :

The first derivative of a cubic is a :

The second derivative of a cubic is a :

line(ar)

constant

quadratic

line(ar)

The second derivative of a quartic (x4) is - quadratic

This pattern should be easy to understand beforewe continue – remember, 1

0 1

N N

n nn n

n n

dyy a x na x

dx

A little quizA little quiz

Look at these curves. Which is the “odd one out”?

AB C

F

E

D

CurvatureCurvature

The answer is D – all other curves have negative curvature

Curvature is measured by the second derivative.Look:

•The gradient is always increasing•The second derivative is always positive•The curvature is positive

2

20

d yx

dx

•The gradient is always decreasing•The second derivative is always negative•The curvature is negative

2

20

d yx

dx

Stationary points revisitedStationary points revisited

We can use the idea of the second derivative to improveour method of identifying maxima and minima.

Let’s return to the cubic equation we looked at earlier:3 23 3 8y x x x

The method given on p. 102 of the textbook is absolutelyaccurate, but can be slow and awkward – because we mustevaluate the first derivative at 2 points.

Look at the graph again, and think about curvature…

Stationary points revisitedStationary points revisited

-20

-15

-10

-5

0

5

10

15

20

-2 -1 0 1 2 3 4cubic

?dy

dx

?dy

dx

2

2?

d yor

dx

Therefore: zero gradient and positive curvature => minimumwhile zero gradient and negative curvature => maximum

The calculation in detailThe calculation in detail

3 23 3 8y x x x 23 6 3 0

dyx x

dx

1 2 or 1 2x 2

26 6 6( 1)

d yx x

dx

1 2x is a minimum, 1 2x is a maximum

which is positive for x>1,negative for x<1

Hence:

at a stationary point

(Don’t forget to calculate the function value!)

Follow the right procedure!Follow the right procedure!

Make sure to read and understand the beginning ofSection 15.3 p. 230. Not all examples are as easy asthe one I’ve just presented!

In particular, think carefully about:•What if f(x) is undefined?•What if f’(x) is undefined?•What if f’’(x) is zero?

The last case, f’’(x) or 2

20

d f

dx is particularly tricky.

Please remember,

f(x) has a stationary point at x=x0

2

2 0d f

dx

Zero curvature and inflexion points

An inflexion point is a point about which the curvature of the function changes sign

Does3 23 3 8y x x x have any inflexion points?

Note – stationary points are not always inflexion points, itdepends on the curvature or second derivative.

Consider the function y=x3 at x=0. 2

20

d yx

dx

2

20

d yx

dx

2

20

d yx

dx

< 0

= 0

> 0

So we see that the curvature changes sign;

first it bends one way, then the other way. This is a Point of Inflexion

Inflexion points continued

What can we say about the tangent at inflexion points?

Answer – the tangent crosses the curve! But this onlyapplies to smooth, differentiable functions. Let’s look atsome “nastier” curves:

First derivative not defined here; but is it an inflexion point?

First derivative not defined here; but is it an inflexion point? Yes!

No!

Note: this type will almost certainly notbe on the P1 exam.

Higher order derivatives

So far we have looked at first and second derivatives, whichare very commonly used in a variety of applications.higher order derivatives are also sometimes useful.it is easy to understand how to find them; just use the sameprocedures.

2 3 4

2 3 4, , , ....

dy d y d y d y

dx dx dx dx

Called “third, fourth,fifth derivative” etc.

Example: find the fourth derivative of 5 4( ) 3f x x x

120 72x