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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