# Implicit Differentiation

*Finding the derivative when you can’t solve for y *

You may like to read Introduction to Derivatives and Derivative Rules first.

## Implicit vs Explicit

A function can be explicit or implicit:

**Explicit**: "y = some function of x". *When we know x we can calculate y directly.*

**Implicit**: "some function of y and x equals something else". *Knowing x does not lead directly to y.*

### Example: A Circle

Explicit Form | Implicit Form | |

y = ± √ (r^{2} − x^{2}) | x^{2} + y^{2} = r^{2} | |

In this form, y is expressed as a function of x. | In this form, the function is expressed in terms of both y and x. |

The graph of x^{2} + y^{2} = 3^{2}

## How to do Implicit Differentiation

- Differentiate with respect to x
- Collect all the \frac{dy}{dx} on one side
- Solve for \frac{dy}{dx}

### Example: x^{2} + y^{2} = r^{2}

Differentiate with respect to x:

\frac{d}{dx}(x^{2}) + \frac{d}{dx}(y^{2}) = \frac{d}{dx}(r^{2})

Let's solve each term:

^{2}) = 2x

^{2}) = 2y\frac{dy}{dx}

^{2}is a constant, so its derivative is 0:\frac{d}{dx}(r

^{2}) = 0

Which gives us:

2x + 2y\frac{dy}{dx} = 0

Collect all the \frac{dy}{dx} on one side

y\frac{dy}{dx} = −x

Solve for \frac{dy}{dx}:

\frac{dy}{dx} = \frac{−x}{y}

### The Chain Rule Using \frac{dy}{dx}

Let's look more closely at how \frac{d}{dx}(y^{2}) becomes 2y\frac{dy}{dx}

The Chain Rule says:

\frac{du}{dx} = \frac{du}{dy}\frac{dy}{dx}

Substitute in u = y^{2}:

\frac{d}{dx}(y^{2}) = \frac{d}{dy}(y^{2})\frac{dy}{dx}

And then:

\frac{d}{dx}(y^{2}) = 2y\frac{dy}{dx}

### Basically, all we did was differentiate with respect to y and multiply by \frac{dy}{dx}

Another common notation is to use ’ to mean \frac{d}{dx}

### The Chain Rule Using ’

The Chain Rule can also be written using ’ notation:

f(g(x))’ = f’(g(x))g’(x)

g(x) is our function "y", so:

f(y)’ = f’(y)y’

f(y) = y^{2}, so f’(y) = 2y:

f(y)’ = 2yy’

or alternatively: f(y)’ = 2y \frac{dy}{dx}

### Again, all we did was differentiate with respect to y and multiply by \frac{dy}{dx}

## Explicit

Let's also find the derivative using the **explicit** form of the equation.

- To solve this explicitly, we can solve the equation for y
- Then differentiate
- Then substitute the equation for y again

### Example: x^{2} + y^{2} = r^{2}

^{2}from both sides:y

^{2}= r

^{2}− x

^{2}

^{2}− x

^{2})

**just the positive**: y = √(r

^{2}− x

^{2})

^{2}− x

^{2})

^{½}

**Derivative (Chain Rule)**:y’ =½(r

^{2}− x

^{2})

^{−½}(−2x)

^{2}− x

^{2})

^{−½}

**y = (r**: y’ = −x/y

^{2}− x^{2})^{½}We get the same result this way!

You can try taking the derivative of the negative term yourself.

### Chain Rule Again!

Yes, we used the Chain Rule again. Like this (note different letters, but same rule):

\frac{dy}{dx} = \frac{dy}{df}\frac{df}{dx}

Substitute in f = (r^{2} − x^{2}):

\frac{d}{dx}(f^{½}) = \frac{d}{df}(f^{½})\frac{d}{dx} (r^{2} − x^{2})

Derivatives:

\frac{d}{dx}(f^{½}) = ½(f^{−½}) (−2x)

And substitute back f = (r^{2} − x^{2}):

\frac{d}{dx}(r^{2} − x^{2})^{½} = ½((r^{2} − x^{2})^{−½}) (−2x)

And we simplified from there.

## Using The Derivative

OK, so why find the derivative y’ = −x/y ?

Well, for example, we can find the slope of a tangent line.

### Example: what is the slope of a circle centered at the origin with a radius of 5 at the point (3,4)?

No problem, just substitute it into our equation:

\frac{dy}{dx} = −x/y

\frac{dy}{dx} = −3/4

And for bonus, the equation for the tangent line is:

y = −3/4 x + 25/4

## Another Example

Sometimes the **implicit** way works where the explicit way is hard or impossible.

### Example: 10x^{4} - 18xy^{2} + 10y^{3} = 48

How do we solve for y? We don't have to!

- First, differentiate with respect to x (use the Product Rule for the xy
^{2}term). - Then move all dy/dx terms to the left side.
- Solve for dy/dx

Like this:

^{4}− 18xy

^{2}+ 10y

^{3}= 48

**Derivative**:10 (4x

^{3}) − 18(x(2y\frac{dy}{dx}) + y

^{2}) + 10(3y

^{2}\frac{dy}{dx}) = 0

(the middle term is explained below)

^{3}− 36xy\frac{dy}{dx} − 18y

^{2}+ 30y

^{2}\frac{dy}{dx} = 0

^{2}\frac{dy}{dx} = −40x

^{3}+ 18y

^{2}

^{2}−36xy)\frac{dy}{dx}= 18y

^{2}− 40x

^{3}

^{2}−6xy)\frac{dy}{dx}= 9y

^{2}− 20x

^{3}

And we get:

\frac{dy}{dx} = | 9y^{2} − 20x^{3} |

3(5y^{2} − 6xy) |

### Product Rule

For the middle term we used the Product Rule: (fg)’ = f g’ + f’ g

^{2})’ = x(y

^{2})’ + (x)’y

^{2}

^{2}

Because (y^{2})’ = 2y\frac{dy}{dx}(we worked that out in a previous example)

Oh, and \frac{dx}{dx} = 1, in other words x’ = 1

## Inverse Functions

Implicit differentiation can help us solve inverse functions.

The general pattern is:

- Start with the inverse equation in explicit form. Example: y = sin
^{−1}(x) - Rewrite it in non-inverse mode: Example: x = sin(y)
- Differentiate this function with respect to x on both sides.
- Solve for dy/dx

As a final step we can try to simplify more by substituting the original equation.

An example will help:

### Example: the inverse sine function y = sin^{−1}(x)

^{−1}(x)

**Derivative**:\frac{d}{dx}(x) = \frac{d}{dx}sin(y)

We can also go one step further using the Pythagorean identity:

sin^{2} y + cos^{2} y = 1

cos y = √(1 − sin^{2} y )

And, because sin(y) = x (from above!), we get:

cos y = √(1 − x^{2})

Which leads to:

\frac{dy}{dx}= \frac{1}{√(1 − x^{2})}

### Example: the derivative of square root √x

^{2}= x

**Derivative**:2y\frac{dy}{dx}= 1

Note: this is the same answer we get using the Power Rule:

^{½}

^{n}= nx

^{n−1}:\frac{dy}{dx} = (½)x

^{−½}

## Summary

- To Implicitly derive a function (useful when a function can't easily be solved for y)
- Differentiate with respect to x
- Collect all the dy/dx on one side
- Solve for dy/dx

- To derive an inverse function, restate it without the inverse then use Implicit differentiation