# Implicit Differentiation

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

## Example 1

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*.

[*x*^{2}]' = 2*x*^{1}

[*y*^{2}]' = 2*y*^{1}⋅*y*'*y*' is multiplied

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

So the chain rule is used:

first differentiate *y*^{2}, 2*y*^{1},

then differentiate *y*, *y*' (= *dy*/*dx*).

And [-1]' = 0.

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

Move the term that doesn't have *y*' (2*x*)

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

Differentiate each term with respect to *x*.

[*x*^{2}]' = 2*x*^{1}

[*xy*^{2}]' = 1*x*^{0}⋅*y*^{2} + *x*⋅2*y*^{1}⋅*y*'

Chain rule in differentiation

[-2*y*^{3}]' = -2⋅3*y*^{2}⋅*y*'

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*'(2*xy* + 6*y*^{2}).

Divide both sides by (2*xy* + 6*y*^{2}).

Then *y*' = (-3*x*^{2} - *y*^{2}) / (2*xy* + 6*y*^{2}).

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

Then [*y*']_{(x, y) = (1, 1)} = -1/2.