Implicit Differentiation

Implicit Differentiation

How to solve the implicit differentiation problems: examples and their solutions.

Example 1

For the given function, find dy/dx. x^2 + y^2 - 1 = 0

An implicit function is a function
that cannot be written as 'y = ...' or 'x = ...'.
(The variables cannot be separated.)

So an implicit function is written as 'f(x, y) = 0'.

Let's see how to differentiate implicit functions.

Differentiate each term with respect to x.

[x2]' = 2x1

[y2]' = 2y1y'

y' is multiplied
because the differentiation is with respect to x, not y.

So the chain rule is used:
first differentiate y2, 2y1,
then differentiate y, y' (= dy/dx).

And [-1]' = 0.

Change the equation into [y' = ...] form.

Move the term that doesn't have y' (2x)
to the right side.

Divide both sides by 2.
And divide both sides by y.

Then y' = -x/y.

This is dy/dx.

Example 2

For the given function, find dy/dx at (1, 1). x^3 + xy^2 - 2y^3 + 2 = 0

Differentiate each term with respect to x.

[x2]' = 2x1

[xy2]' = 1x0y2 + x⋅2y1y'

Chain rule in differentiation

[-2y3]' = -2⋅3y2y'

And [+2]' = 0.

Change the equation into [y' = ...] form.

Move the terms that don't have y'
to the right side.

And factor the remaining terms using y':
y'(2xy + 6y2).

Divide both sides by (2xy + 6y2).

Then y' = (-3x2 - y2) / (2xy + 6y2).

Put (1, 1) into (x, y).

Then [y'](x, y) = (1, 1) = -1/2.