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Techniques for differentiation

Section 8: Implicit differentiation

Notes and Examples

These notes contain subsections on

Differentiating functions of y with respect to x Differentiating implicit functions

Differentiating products of x and y Logarithmic differentiation

Differentiating functions of y with respect to x

Suppose that y is a function of x, say 2x . You can differentiate y to get 2x; but

what about differentiating 2y , or sin y or ln y ? You could convert them to

functions of x, i.e. 4 2 2, sin lnx x xor ; but can you find the derivative of these

functions without converting into functions of x?

For example, suppose 2y x as above and you want to find the derivative of

y² with respect to x. Let 2u y . You want to findd

d

u

x

2 d2

d

uu y y

y

2 d2

d

yy x x

x .

Using the chain rule:

2

3

d d d

d d d

d2

d

2 2

4

u u y

x y x

yy

x

x x

x

(which is indeed the derivative of x4)

The other functions of y can be differentiated in the same way:

To differentiate sin y with respect to x, let sinu y . You want to findd

d

u

x.

dsin cos

d

uu y y

y

2 d2

d

yy x x

x

Using the chain rule:

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2d d[ ] 2

d d

yy y

x x , as before.

d d d

d d d

dcos

d

2 cos

u u y

x y x

yy

x

x y

To differentiate ln y with respect to x, let lnu y . You want to findd

d

u

x.

d 1ln

d

uu y

y y

2 d2

d

yy x x

x

Using the chain rule:

2

d d d

d d d

1 d

d

12

2

u u y

x y x

y

y x

xx

x

This idea can be generalised:

If y is a function of x, then d df d[f ( )]

d d d

yy

x y x .

Differentiating implicit functions

An implicit function is an equation between two variables, say x and y, inwhich y is not given explicitly as a function of x. The equation of a circle is anexample of an implicit function.

For example, 2 2 25x y is the equation of a circle, centre O and radius 5.

You can differentiate an implicit function like this by differentiating each side ofthe equation with respect to x:

2 2 25x y

2 2d d d[ ] [ ] [25]

d d dx y

x x x

d2 2 0

d

yx y

x

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You can now findd

d

y

xin terms of x and y by rearranging this equation:

d2 2 0

d

yx y

x

d2 2

d

yy x

x

d 2

d 2

y x x

x y y

This result enables you, for example, to find the gradient at the point (3, 4) of

the circle. Here, x = 3 and y = 4, so d 3

d 4

y x

x y .

Example 1

Show that the curve y + y3 = 2sin x has a turning point at the point 2 ,1 .

Solution3 2siny y x

Differentiating implicitly gives:

3d d[ ] [2sin ]

d dy y x

x x

2d d3 2cos

d d

y yy x

x x ,

2d(1 3 ) 2cos

d

yy x

x

2

d 2cos

d 1 3

y x

x y

At the point 2 ,1 x = 2 and y = 1

22cosd0

d 1 3

y

x

So 2 ,1 is a turning point.

Differentiating products of x and y

What is the derivative of 2xy with respect to x?

You can use the product rule to do this.d

1d

uu x

x and 2 d d

2d d

v yv y y

x x .

So 2d d d( )

d d d

d2

d

v uxy u v

x x x

yxy y

x

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Example 2

Given that y2 + 3xy + x3 = 25, finddy

dxin terms of x and y.

Solution2 33 25y xy x

Differentiating implicitly gives:

2 3d d[ 3 ] [25]

d dy xy x

x x

2d d2 3 3 3 0

d d

y yy x y x

x x

2d(2 3 ) 3 3

d

yy x y x

x

2

d (2 3 )

d 3 3

y y x

x y x

You may also like to look at the Implicit differentiation video.

Logarithmic differentiation

The technique of implicit differentiation can be used to simplify differentiatingcomplicated product and quotient functions, as in the following example.

Example 3

Find the derivative of3

2

(1 )(1 2 )

1

x xy

x

Solution3

2

(1 )(1 2 )

1

x xy

x

Take logarithms of each side:3

2

3 2

2

(1 )(1 2 )ln ln

1

ln(1 ) ln(1 2 ) ln 1

1ln(1 ) 3ln(1 2 ) ln(1 )

2

x xy

x

x x x

x x x

Now differentiate implicitly:

2

1 d 1 2 1 2( 2)

d 1 1 2 2 (1 )

y x

y x x x x

The product rule is used here to

differentiate 3xy, with u = 3x and v = y.It is not necessary to write out all thedetails of the product rule if you don’tfeel that you need to.

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2

3

22

d 1 4

d 1 1 2 (1 )

(1 )(1 2 ) 1 4

1 1 2 (1 )1

y xy

x x x x

x x x

x x xx

For some practice in identifying the correct differentiation technique to beused, try the Mathsnet resource Method match.

This is quite a fearsome looking expression! But try

differentiating y directly and you will appreciate thatthe logarithmic differentiation is considerably easier!