38: Implicit Differentiation © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

38: Implicit 38: Implicit DifferentiationDifferentiation

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

Implicit Differentiation

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Module C3 AQA

Edexcel

Module C4

MEI/OCR

OCR


We usually write a relationship between x and y in the form

)(xfy

However, we’ve already met some curves, for example, a circle, where it is easier to have x and y “mixed up” on the same side of the equation.

This gives y explicitly in terms of x.

e.g. 342 xxy is explicit

422 yx is implicit

These give y implicitly.


It’s not always easy to rearrange implicit formulae to explicit form so we need to be able to differentiate them as they are.

2xy

We’ll first think about differentiating

( explicit )

This is called differentiating with respect to (w.r.t.) xSo, differentiating y w.r.t. x gives

We get x2dx

dy


2xy


( explicit )

This is called differentiating with respect to (w.r.t.) xSo, differentiating y w.r.t. x gives dx

dy

Also, differentiating with respect to x is easy.

2x

We get x2dx

dy

It’s not always easy to rearrange implicit formulae to explicit form so we need to be able to differentiate them as they are.


Also, differentiating with respect to x is easy.

2x

It isn’t usually easy to rearrange implicit formulae to explicit form so we need to be able to differentiate them as they are.

2xy


( explicit )

We get x2This is called differentiating with respect to (w.r.t.) x

So, differentiating with respect to y is also easy.

2yIt gives 2y.

So, differentiating y w.r.t. x gives dx

dy

dx

dy


dy

dy

dx

)( 2yd

dx

yd )( 2

dx

du

du

dy

dx

dy

To differentiate with respect to x we use the chain rule.

2y

The chain rule for differentiating :

2y

becomes

We choose y as the “chaining variable” here . . .

Differentiating with respect to y gives 2y.

2y


dx

dydx

du

du

dy

dx

dy


2y


2y

becomes



2y

So,dx

dyy

dx

yd2

)( 2

ordx

dyy

dx

yd2

)( 2

ydy

yd2

)( 2

because we’ve just seen that

dy

)( 2yd

dx

yd )( 2


We can now differentiate the following:

dx

dyy2w.r.t. x, giving

2y

and as

b

e

f

o

r

e

:

w.r.t. x, giving2x x2

dx

dyy w.r.t. x, givingand

Tip: Whenever we have a function of y to differentiate w.r.t. x, we just differentiate it, using the usual rules,

dx

dyand multiply by .

Implicit Differentiatione.g. 1 Differentiate the following with

respect to x422 yx(a)

(b) 123 32 yxxy

Solution: (a) 422 yx

x2 dx

dyy2 0

(b) 123 32 yxxy

dx

dy1 x6 26 y

dx

dy0


Solutio

n

:

3

2x0

e.g. 2 A curve is defined by the equation

143

22

yx

Find the gradient at the point . )1,( 23

143

22

yx

4

2 y

dx

dy

Tip: Don’t rearrange to

find unless you are

asked to. Just

substitute.

dx

dy

1,23 yx

04

2

3

)(2 23

m1

2

01 21 m

2 m121 m


Solutio

n

:

e.g. 3 Find the gradient function of the curve

21ln xy giving your answer in the

form .),( yxfdx

dy

21ln xy

xdx

dy

y2

1

xydx

dy2

This means the function on the r.h.s. can contain x and y.

Implicit DifferentiationExercise1. Differentiate the following with respect to

x:233 22 yxyx(a)

(b)23 xy

2. Find the gradient at ( 1, 3 ) on the curve

04422 xyx


(b)23 xy

2. Find the gradient at ( 1, 3 ) on the curve04422 xyx

233 22 yxyx1(a

)Solutio

n

:

02233 dx

dyyx

dx

dy

xdx

dyy 23 2

Solutio

n

:0422

dx

dyyx

:3,1 yxSubst.

0462 m 1 m

Implicit DifferentiationWe may have to differentiate terms such as

xy.

dx

xyd )(

This is a product so we use the product rule:

1ydx

dyx

dx

dvu

dx

duv

dx

uvd

)( become

s

dx

dyxy

Tip: With implicit equations I always look for a product and if I see one I write P by it so

that I don’t then forget it !

dx

xyd )( dx

dxy

dx

dyx


Solutio

n

:

x2dx

dyy2

dx

dyxy 0

P

Can you see how we could make a mistake with this product?

ANS: The minus sign. There will be 2 terms so we must use brackets.

222 xyyxSo,

e.g. 1 Given show that 222 xyyxxy

xy

dx

dy

2

2


xydx

dyx

dx

dyy 22

x2dx

dyy2

dx

dyxy 0

In this question, we do have to rearrange to

find dx

dy

Collect terms containing on the l.h.s. and

the others on the r.h.s. dx

dy

xyxydx

dy2)2(

xy

xy

dx

dy

2

2

Common

factor:

Divide by ( 2y x ):

Implicit Differentiatione.g. 2 The equation of a curve

is 2823 22 xxyyx

Solutio

n

:

Find an equation connecting x, y and at all

points on the curve. Hence show that the

coordinates of the points on the curve at which

satisfy the equation 4 yx

dx

dy

2dx

dy

2823 22 xxyyxP

x6 8dx

dyy2

dx

dyxy2

2dx

dy 8)2(246 xyyx

822 yx 4 yx


Solutio

n

:

e.g. 3 Find an equation linking x, y and

if1443 322 xxyyx

dx

dy

x2

P

dx

dyy6 34 y 4

dx

dyyx 23

412462 23 dx

dyxyy

dx

dyyx

Implicit DifferentiationExercise

2. Find the gradients at the 2 points on the

curve given by 285 22 xxyy

where .

1. A curve is given by the

equation0824 22 yxyx

1x( You will need to find the values of y at x = 1.

)

Find an equation linking x, y and and

find the value of at the point ( 0, 2 ).dx

dy dx

dy


1. 0824 22 yxyxSolutions:

0442

dx

dyy

dx

dyxyx

0)2(4)2(4 m 2,0 yx1m

The gradient at (0, 2) is

1.

P


016254 mm 32 yy or

:)2(2,1 in yx0165104 mm

:)2(3,1 in yx 016356 mm0165156 mm

1m

2.

285 22 xxyy )1(

01652

x

dx

dyxy

dx

dyy )2(

:)1(1 inx 2852 yy 0652 yy 0)3)(2( yy

6m

The gradients are 6 and

1

P

Implicit DifferentiationExercise3. A curve is given by the

equation03353 222 yxyx

Find an equation linking x, y and

. dx

dy

Solutio

n

:

03353 222 yxyxP

06256 2

dx

dyy

dx

dyyxyx

061056 2 dx

dyy

dx

dyxyyx


One useful application of implicit differentiation may arise in growth and decay problems which we look at later.

xy 2lnln 2lnln xy

dx

dy

2lnydx

dy

xy 2e.g. Suppose we want to find given thatdx

dy

We can’t differentiate when x is an index ( apart from ), so we have to take logs.xeUsing base e: xy 2Using the 3rd law of logs:

2lnDifferentiate w.r.t. x:y

1

is just a constant

2ln

Substitute for y:

2ln2 x

dx

dy


xay

Generalising the last result:

aadx

dy x ln



The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.


We usually write a relationship between x and y in the form

)(xfy

However, we’ve already met some curves, for example, a circle, where it is easier to have x and y “mixed up” on the same side of the equation.

This gives y explicitly in terms of x.

e.g. 342 xxy is explicit

422 yx is implicit

These give y implicitly.


dy

dxdx

du

du

dy

dx

dy


2y


2y

becomes



2y

So,dx

dyy

dx

yd2

)( 2

ordx

dyy

dx

yd2

)( 2

ydy

yd2

)( 2

because we’ve just seen that

dy

)( 2yd

dx

yd )( 2


We can now differentiate the following:

dx

dyy2w.r.t. x, giving

2y

and as

b

e

f

o

r

e

:

w.r.t. x, giving2x x2

dx

dyy w.r.t. x, givingand

Tip: Whenever we have a function of y to differentiate w.r.t. x, we just differentiate it, using the usual rules,

dx

dyand multiply by .


Solutio

n

:

3

2x0

e.g. A curve is defined by the equation

143

22

yx

Find the gradient at the point . )1,( 23

143

22

yx

4

2 y

dx

dy

Tip: Don’t rearrange to

find unless you are

asked to. Just

substitute.

dx

dy

1,23 yx

04

2

3

)(2 23

m1

2

01 21 m

2 m121 m

Implicit DifferentiationWe may have to differentiate terms such as

xy.

dx

xyd )(

This is a product so we use the product rule:

1ydx

dyx

dx

dvu

dx

duv

dx

uvd

)( become

s

dx

dyxy

Tip: With implicit equations I always look for a product and if I see one I write P by it so

that I don’t then forget it !

dx

xyd )( dx

dxy

dx

dyx

Implicit Differentiatione.g. The equation of a curve

is 2823 22 xxyyx

Solutio

n

:

Find an equation connecting x, y and at all

points on the curve. Hence show that the

coordinates of the points on the curve at which

satisfy the equation 4 yx

dx

dy

2dx

dy

2823 22 xxyyxP

x6 8dx

dyy2

dx

dyxy2

2dx

dy 8)2(246 xyyx

822 yx 4 yx

Documents

38: Implicit Differentiation © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules