# Derivative of an Implicit Function

How to find the derivative of an implicit function: 2 examples and their solutions.

## Example 1

### Example

An implicit function is a function that looks like f(x, y) = 0.

It cannot be written as x = ... or y = ... .

Let's see how to find the derivative of an implicit function.

Actually, x^{2} + y^{2} = 1 is a circle.

So it's not a [function].

But in this chapter,

we use the term function as an equation.

### Solution

See y as a variable

and write the derivative of y^{2}: 2y^{1}.

And write the derivative of y: y'.

It's like the derivative of a composite function:

First differentiate the outer function y^{2}: 2y^{1}.

Then differentiate the inner function y: y'.

For this reason,

when differentiating a term with y,

multiply y'.

The right side is a constant.

So write = 0.

Then you get 2x^{1} + 2y^{1}⋅y' = 0.

Write this equation as y' = ... .

2x^{1} = 2x

+2y^{1}⋅y' = +2yy'

Divide both sides by 2.

Move x to the right side.

Divide both sides by y.

Then y' = -x/y.

So dy/dx = y' = -x/y.

## Example 2

### Example

### Solution

Differentiate the given function.

Write the derivative of x^{3}: 3x^{2}.

Write the derivative of +xy^{2}.

Write,

the derivative of x, 1 times y^{2}

plus

x times, the derivative of y^{2}, 2y^{1}⋅y'.

Derivative of a Product

Don't forget to multiply y'

when writing the derivative of y^{2}: 2y^{1}⋅y'.

Write the derivative of -2y^{3}:

-2⋅3y^{2}⋅y'.

Don't forget to write y'.

+2 is a constant.

So write 0.

The right side is 0.

So write = 0.

So 3x^{2} + 1⋅y^{2} + x⋅2y^{1}⋅y' - 2⋅3y^{2}⋅y' + 0 = 0.

Write this equation as y' = ... .

+1⋅y^{2} = +y^{2}

+x⋅2y^{1}⋅y' = +2xyy'

-2⋅3y^{2}⋅y' = -6y^{2}y'

Move 3x^{2} + y^{2} to the right side.

And combine +2xyy' - 6y^{2}y':

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

Common Monomial Factor

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

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

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

This is dy/dx.

It says to find dy/dx at (1, 1).

So find [y']_{(x, y) = (1, 1)}.

Put (x, y) = (1, 1) into y' = (-3x^{2} - y^{2})/(2xy - 6y^{2}).

Then [y']_{(x, y) = (1, 1)} = (-3⋅1^{2} - 1^{2})/(2⋅1⋅1 - 6⋅1^{2}).

Then you get 1.

So dy/dx = y' = 1 at (1, 1).