# 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, x2 + y2 = 1 is a circle.
So it's not a [function].
But in this chapter,
we use the term function as an equation.

### Solution

Differentiate the given function.

Write the derivative of x2: 2x1.

Derivative of a Polynomial

See y as a variable
and write the derivative of y2: 2y1.
And write the derivative of y: y'.

It's like the derivative of a composite function:
First differentiate the outer function y2: 2y1.
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 2x1 + 2y1⋅y' = 0.

Write this equation as y' = ... .

2x1 = 2x
+2y1⋅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

### Solution

Differentiate the given function.

Write the derivative of x3: 3x2.

Write the derivative of +xy2.

Write,
the derivative of x, 1 times y2
plus
x times, the derivative of y2, 2y1⋅y'.

Derivative of a Product

Don't forget to multiply y'
when writing the derivative of y2: 2y1⋅y'.

Write the derivative of -2y3:
-2⋅3y2⋅y'.

Don't forget to write y'.

+2 is a constant.
So write 0.

The right side is 0.
So write = 0.

So 3x2 + 1⋅y2 + x⋅2y1⋅y' - 2⋅3y2⋅y' + 0 = 0.

Write this equation as y' = ... .

+1⋅y2 = +y2
+x⋅2y1⋅y' = +2xyy'
-2⋅3y2⋅y' = -6y2y'

Move 3x2 + y2 to the right side.

And combine +2xyy' - 6y2y':
+y'(2xy - 6y2).

Common Monomial Factor

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

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

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

Then [y'](x, y) = (1, 1) = (-3⋅12 - 12)/(2⋅1⋅1 - 6⋅12).

Then you get 1.

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