c3.6 Differentiation

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A-Level Maths: Core 3for Edexcel

C3.6 Differentiation

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© Boardworks Ltd 20062 of 56

The chain rule

The relationship between

Differentiating ex and related functions

Differentiating ln x and related functions

The product rule

The quotient rule

Differentiating trigonometric functions

Examination-style question

dy dxdx dyand

Co

nte

nts


The chain rule


Review of differentiation

So far, we have used differentiation to find the gradients of functions made up of a sum of multiples of powers of x. We found that:

and when xn is preceded by a constant multiplier k we have:

1If = then =n ndyy kx knx

dx

f g f gIf = ( ) ± ( ) then = '( ) ± '( )dy

y x x x xdx

Also:

1If = then =n ndyy x nx

dx


Review of differentiation

We will now look at how to differentiate exponential, logarithmic and trigonometric functions.

We will also look at techniques that can be used to differentiate:

Compound functions of the form f(g(x)). For example:

2 1x xe 3 2 x3sin( )

Products of the form f(x) × g(x), such as:

x xln xxe 2 x x23 cos

x

x 2

3 +1

1

xe

x

2

sin

x

x

2

ln

Quotients of the form , such as:f

g

( )

( )

x

x


The chain rule

The chain rule is used to differentiate composite functions.

For instance, suppose we want to differentiate y = (2x + 1)3 with respect to x.

One way to do this is to expand (2x + 1)3 and differentiate it term by term.

Using the binomial theorem:

x x x x3 3 2(2 +1) = 8 + 3(4 )+ 3(2 )+1

x x x3 2= 8 +12 + 6 +1

dyx x

dx2= 24 + 24 + 6

x x2= 6(4 + 4 +1)

x 2= 6(2 +1)

Differentiating with respect to x:


The chain rule

Another approach is to use the substitution u = 2x + 1 so that we can write y = (2x + 1)3 as y = u3.

The chain rule states that:

If y is a function of u and u is a function of x, thendy dy du

dx du dx= ×

So if y = u3 where u = 2x + 1,dy

udu

2= 3du

dx= 2

Using the chain rule:dy dy du

dx du dx= × u2= 3 ×2

u2= 6

x 2= 6(2 +1)


The chain rule

121

2=dy

udu

dux

dx= 6

Using the chain rule,dy dy du

dx du dx= ×

121

2= ×6u x

122= 3 (3 5)x x

Use the chain rule to differentiate y = with respect to x.x 23 5

Let where u = 3x2 – 5y u12=

2

3=

3 5

x

x

xu 12= 3


The chain rule

dyu

du 5= 8

dux

dx 2= 3


dx du dx= × u x 5 2= 8 × 3

x x 2 3 5= 24 (7 )

Let yu4

2=

x

x

2

3 5

24=

(7 )

x u 2 5= 24

Find given that .yx 3 4

2=

(7 )

dy

dx

u 4= 2 where u = 7 – x3


The chain rule using function notation

With practice some of the steps in the chain rule can be done mentally.Suppose we have a composite function

y = g(f(x))

If we let y = g(u) where u = f(x)

then andg= '( )dy

udu

f= '( )du

xdx


dx du dx= × g f= '( )× '( )u x

But u = f(x) so

If y = g(f(x)) then g f f= '( ( ))× '( )dy

x xdx


The chain rule

All of the composite functions we have looked at so far have been of the form y = (f(x))n.

In general, using the chain rule,

If y = (f (x))n then f f1= ( ( )) × '( )ndyn x x

dx

If we use to represent f (x) and to represent f ’(x) we can write this rule more visually as:

1= = n ndyy n

dy

Find the equation of the tangent to the curve y = (x4 – 3)3 at the point (1, –8).


The chain rule

When x = 1,dy

dx 2=12(1 3) = 48

Using y – y1 = m(x – x1) the equation of the tangent at the point (1, –8) is:

y + 8 = 48(x – 1)

y = 48x – 48 – 8

y = 48x – 56

y = 8(6x – 7)

dy dx x

dx dx 4 2 4= 3( 3) × ( 3)

x x4 2 3= 3( 3) × 4

x x 3 4 2=12 ( 3)

y = (x4 – 3)3


The chain rule




The product rule

The quotient rule



dy dxdx dyand

Co

nte

nts


The relationship between dy dxdx dyand



Suppose we are given x as a function of y instead of y as a function of x. For instance,

dy dxdx dyand

x = 4y2

We can find by differentiating with respect to y:dxdy

= 8dx

ydy

Using the chain rule we can write

× =1,dy dx

dx dy

1=

dxdy

dy

dx

So by the above result, if 8y then=dx

dy1

=8

dy

dx y

from which we get:



Find the gradient of the curve with equation x = 2y3 – 3y – 7 at the point (3, 2).

dy dxdx dyand

x = 2y3 – 3y – 7

dxy

dy2= 6 3

At the point (3, 2), y = 2:

dx

dy2= 6(2) 3 = 21

We can now find the gradient using the fact that 1

=dxdy

dy

dx

dy

dx

1=

21


Differentiating inverse functions

= cosdx

ydy

Find , writing your answer in terms of x.1(sin )

dx

dy

Let y = sin–1 x so

The result is particularly useful for differentiating

inverse functions. For example:

1=

dxdy

dy

dx

x = sin y

Using the identity cos2y = 1 – sin2y

But sin y = x so

1=

cos

dy

dx y 2

1=

1 sin y

1

2

1(sin ) =

1

dx

dy x


Co

nte

nts



The chain rule




The product rule

The quotient rule



dy dxdx dyand


The derivative of ex

From this, it follows that

x xdyy e e

dxIf = then =

x xdyy ke ke

dxIf = then =

A special property of the exponential function ex is that

where k is a constant.

For example, if y = 4ex – x3

2= 4 3xdye x

dx


Functions of the form ekx

Suppose we are asked to differentiate a function of the form ekx, where k is a constant. For example,

Differentiate y = e5x with respect to x.

udye

du= = 5

du

dx


dx du dx= × = ×5ue

5= 5 xe

Let where u = 5xuy e=

= 5 ue

In practice, we wouldn’t need to include this much working.


Functions of the form ekx

We would just remember that in general,

For example,

We can use the chain rule to extend this to any function of the form ef(x).

If = then =kx kxdyy e ke

dx

2 2( ) = 2x xde e

dx

7 7( ) = 7x xde e

dx

33( ) =

3

xxd e

edx


Functions of the form ef(x)

If y = ef(x) then we can let


dx du dx= × f= × '( )ue x

Let where u = f(x)uy e=

ff ( )= '( ) xx e

f= '( )du

xdx

udye

du=then

So in general, f ff( ) ( )If = then = '( )x xdyy e x e

dx

In words, to differentiate an expression of the form y = ef(x) we multiply it by the derivative of the exponent.


Functions of the form ef(x)

For example,

Using to represent f(x) and to represent f ’(x):

= = dy

y e edx

3( ) =xdye

dx 3 xe

5 4( ) =xdye

dx 5 45 xe

2 9(5 ) =xdye

dx 2 95×2 =xxe 2 910 xxe


The chain rule




The product rule

The quotient rule



dy dxdx dyand

Co

nte

nts




The derivative of ln x

Remember, ln x is the inverse of ex.

So, if y = ln x

then x = ey

Differentiating with respect to y gives:

ydxe

dy=

1 1= =

ydxdy

dy

dx e

dy

dx x

1=But ey = x so,

dyy x

dx x

1If = ln then =


Functions of the form ln kx

Suppose we want to differentiate a function of the form ln kx, where k is a constant. For example:

Differentiate y = ln 3x with respect to x.

1=

dy

du u= 3

du

dx


dx du dx= ×

3=

u

1=

x

Let where u = 3x= lny u

3=

3x


Functions of the form ln kx

When functions of the form ln kx are differentiated, the k’s will always cancel out, so in general,

We can use the chain rule to extend to functions of the more general form y = ln f(x).

1If = ln then =

dyy kx

dx x


dx du dx= ×

f '( )=

x

u

Let where u = f(x)= lny u

f

f

'( )=

( )

x

x

f= '( )du

xdx

1=

dy

du uthen


Functions of the form ln (f(x))

ff

f

'( )If = ln ( ) then =

( )

dy xy x

dx x

In general, using the chain rule


For example, ln(7 4) =d

xdx

7

7 4x

= ln =

dyy

dy

3ln(3 + 8) =d

xdx

2

3

9

3 + 8

x

x



In some cases we can use the laws of logarithms to simplify a logarithmic function before differentiating it.

Remember that,

ln (ab) = ln a + ln b ln (ab) = ln a + ln b ln = ln lna

a bb

ln an = n ln a ln an = n ln a

Differentiate with respect to x.= ln2

xy

= ln2

xy = ln ln2x

12= ln ln2x

12= ln ln2x



ln 2 is a constant and so it disappears when we differentiate.

12= ln ln2 =

dyy x

dx

1 1= ×

2 x

1=

2x

If we had tried to differentiate without simplifying it first, we would have had:

= ln2

xy

121

2= ln =dy

y xdx

12

12

1412

x

x

1 12 2

1=

2x x1

=2x

The derivative is the same, but the algebra is more difficult.


The chain rule




The product rule

The quotient rule



dy dxdx dyand

Co

nte

nts


The product rule


The product rule

The product rule allows us to differentiate the product of two functions.

It states that if y = uv, where u and v are functions of x, then

dy dv duu v

dx dx dx= +

So3= 4

dux

dx

12

1= (3 2 ) × 2

2

dvx

dx

4Find given that = 3 2 .dy

y x xdx

Let u = x4 and v 12= (3 2 )x

12= (3 2 )x


The product rule

Using the product rule:1 12 24 3= (3 2 ) + 4 (3 2 )

dyx x x x

dx

1 12 2

1 12 2

4 34 (3 2 ) (3 2 )= +

(3 2 ) (3 2 )

x x x x

x x

12

4 3+ 4 (3 2 )=

(3 2 )

x x x

x

12

4 3 4+12 8=

(3 2 )

x x x

x

12

3 412 9=

(3 2 )

x x

x

3 43(4 3 )

=3 2

x x

x


The product rule

Give the coordinates of any stationary points on the curve y = x2e2x.

So = 2du

xdx

2= 2 xdve

dx

Let u = x2 and v 2= xe

Using the product rule:

2 2 2= 2 + 2x xdyx e xe

dx2= 2 ( +1)xxe x

2= 0 when 2 = 0 or +1= 0xdyxe x

dx22 = 0 = 0xxe x

+1= 0 = 1x x


The product rule

When x = 0, y = (0)2e0

= 0

The point (0,0) is a stationary point on the curve y = x2e2x.

When x = –1, y = (–1)2e–2

= e–2

The point (–1, e–2) is also a stationary point on the curve y = x2e2x.


The chain rule




The product rule

The quotient rule



dy dxdx dyand

Co

nte

nts


The quotient rule


The quotient rule

The quotient rule allows us to differentiate the quotient of two functions.

2=

du dvdx dxv udy

dx v

It states that if y = , where u and v are functions of x, thenuv

Find given that .x

yx2

2 +1=

5

dy

dx

Let u = 2x + 1 and v = 5x2

Sodu

dx= 2

dvx

dx=10


The quotient rule

dy x x x

dx x

2

4

(5 )(2) (2 +1)(10 )=

25

x x x

x

2 2

4

10 20 10=

25

x x

x

3

2 4 2=

5

x

x

3

2 2=

5

x

x

3

2( +1)=

5


The quotient rule

Let u = ln x4 and v = x2

So3

4

4=

du x

dx x= 2

dvx

dx

Using the quotient rule:2 1 4

4

× 4 ln ×2=

dy x x x x

dx x

4

4 8 ln=

x x x

x

Find the equation of the tangent to the

curve y = at the point (1, 0).4

2

ln x

x

1= 4x

3

4(1 2ln )=

x

x

Using ln x4 = 4 ln x


The quotient rule

When x = 1,

The gradient of the tangent at the point (1, 0) is 4.

Use y – y1 = m(x – x1) to find the equation of the tangent at the point (1, 0).

y – 0 = 4(x – 1)

y = 4x – 4

4(1 2 ln 1)=

1

dy

dx

= 4

Remember that ln 1 = 0


The chain rule




The product rule

The quotient rule



dy dxdx dyand

Co

nte

nts




The derivative of sin x


The derivative of sin x

By plotting the gradient function of y = sin x, where x is measured in radians, we can deduce that

dyy x x

dxIf = sin then = cos

Functions of the form k sin f(x) can be differentiated using the chain rule.

Differentiate y = 2 sin 3x with respect to x.

So if y = 2 sin u where u = 3x

= 2cosdy

udu

= 3du

dx


The derivative of sin f(x)


= sin = cosdy

ydy


dx du dx= × = 2cos ×3u

= 6cos3x

In general using the chain rule,

f f fIf = sin ( ) then = '( )cos ( )dy

y x x xdx


The derivative of cos x



By plotting the gradient function of y = cos x, where x is measured in radians, we can deduce that

dyy x x

dxIf = cos then = sin

Find given that y = –x2 cos x.dy

dx

Let u = –x2 and v = cos x

So = 2du

xdx

= sindv

xdx



Using the product rule:

2= ( sin )+ cos ( 2 )dy

x x x xdx

2= sin 2 cosx x x x

= ( sin 2cos )x x x x

Functions of the form k cos f(x) can be differentiated using the chain rule.

Differentiate y = 3 cos (x3 – 4) with respect to x.

So if y = 3 cos u where u = x3 – 4

= 3sindy

udu

2= 3du

xdx




= cos = sindy

ydy


dx du dx= × 2= 3sin ×3u x

2 3= 9 sin( 4)x x

In general using the chain rule,

f f fIf = cos ( ) then = '( )sin ( )dy

y x x xdx


The derivative of tan x

We can differentiate y = tan x (where x is in radians) by writing it as x

yx

sin=

cosThen we apply the quotient rule with u = sin x and v = cos x :

dy x x x x

dx x

2

cos cos sin ( sin )=

cos

x x

x

2 2

2

cos + sin=

cos

x2

1=

cos

x2= sec

dyy x x

dx2If = tan then = sec


The derivative of sec x

We can differentiate y = sec x (where x is in radians) by writing it as

y xx

11= = (cos )

cos

Then using the chain rule we get:dy

x xdx

2= (cos ) ( sin )

x

x2

sin=

cos

x

x x

1 sin= ×

cos cos

x x= sec tan

dyy x x x

dxIf = sec then = sec tan


The derivative of cosec x

We can differentiate y = cosec x (where x is in radians) by writing it as

y xx

11= = (sin )

sin

Then using the chain rule we get:dy

x xdx

2= (sin ) (cos )

x

x

2

cos=

sin

x

x x

1 cos= ×

sin sinx x= cosec cot

dyy x x x

dxIf = cosec then = cosec cot


The derivative of cot x

We can differentiate y = cot x (where x is in radians) by writing it as x

yx

cos=

sinThen we apply the quotient rule with u = cos x and v = sin x:

dy x x x x

dx x

2

sin ( sin ) cos cos=

sin

x x

x

2 2

2

(sin + cos )=

sin

x

2

1=

sin

x 2= cosec

2If = cot then = cosec dy

y x xdx


dyy x x

dx= sin = cos

Derivatives of trigonometric functions

In summary, if x is measured in radians, then

2= cot = cosec dy

y x xdx

dyy x x x

dx= sec = sec tan

dyy x x x

dx = cosec = cosec cot

dyy x x

dx = cos = sin

dyy x x

dx 2= tan = sec

When learning these results, it is helpful to notice that all of the trigonometric functions starting with ‘co’ have negative derivatives.


The chain rule




The product rule

The quotient rule



dy dxdx dyand

Co

nte

nts





a) find f ’(x),

b) find the coordinates of any stationary points and determine their nature,

c) sketch the curve y = f(x).

Given that , f2

2( ) =

+ 4

xx

x

a) Using the quotient rule: f 2'( ) =

du dvdx dxv u

xv

f2

2 2

2( + 4) 2 (2 )'( ) =

( + 4)

x x xx

x

x x

x

2 2

2 2

2 + 8 4=

( + 4)

2

2 2

2(4 )=

( + 4)

x

x



b) When f(x) = 0,x

x

2

2 2

2(4 )= 0

( + 4)

x 22(4 ) = 0

x 24 = 0

x = 2

When x = 2, y4

= 8

1=

2

When x = –2, y 4

= 8

1

= 2

Therefore, the graph of the function has turning points at (2, ) and (–2, – ).

xx

x2

2f( ) =

+ 412

12



Looking at the gradient just before and just after x = 2:

12So (2, ) is a maximum point.

x 1.9 2 2.1

Value of2

2 2

2(4 )=

( + 4)

dy x

dx x

Slope

Looking at the gradient just before and just after x = –2:

12So (–2, – ) is a minimum point.

x –2.1 –2 –1.9

Value of2

2 2

2(4 )=

( + 4)

dy x

dx x

Slope

–ive0

+ive

0.01 0 –0.01

0.01 0 –0.01

–ive 0 +ive



When x = 0,

c) The curve crosses the axes when x = 0 and when y = 0.

2

2=

+ 4

xy

x

y = 0. (Also, when y = 0, x = 0).

Therefore the curve has one crossing point at the origin, a minimum at (–2, – ) and a maximum at (2, ):1

212

Also,

as , x y 0–

0 and,

as , x y 0+.

0

2

2=

+ 4

xy

x

0–2 212

12

x

y

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