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 No.1[Reply]

• differentiate functions defined implicitly

Contents

1. Introduction 2

2. Revision of the chain rule 2

3. Implicit differentiation with respect to y and then

multiply by dx dy .

Key Point

d

dx (f(y)) = dy d (f(y)) × dx dy

We are now ready to do some implicit differentiation. Remember, every time we want to differentiate a function of y with respect to x, we differentiate with respect to y and then multiply by

dy

dx.

www.mathcentre.ac.uk 3 c mathcentre 20093. Implicit differentiation

Example

Suppose we want to differentiate the implicit function

y2 + x3 − y3 + 6 = 3y

with respect x.

We differentiate each term with respect to x:

d

dx y2 + dx d x3 − dx d y3 + dx d (6) = dx d (3y)

Differentiating functions of x with respect to x is straightforward. But when differentiating a

function of y with respect to x we must remember the rule given in the previous keypoint. We

find

d

dy

y2 × dx dy + 3x2 − dy d y3 × dx dy + 0 = dy d (3y) × dx dy

that is

2y

dy

dx + 3x2 − 3y2dx dy = 3dx dy

We rearrange this to collect all terms involving dx dy together.

3x2 = 3dy

dx − 2y

dy

dx + 3y2dx dy

then

3x2 = 3 − 2y + 3y2 dx dy

so that, finally,

dy

dx =

3x2

3 − 2y + 3y2

This is our expression for dy

dx.

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