# Chain Rule in Differentiation

How to solve the chain rule in differentiation problems: formula, proof, examples, and their solutions.

## Formula

The chain rule is a way

to differentiate a composite functioncomposite function: *f*(*g*(*x*)).

[*f*(*g*(*x*))]' = *f*'(*g*(*x*))⋅*g*(*x*)

First differentiate the outer function

without touching the inner function:*f*'(*g*(*x*)).

Then differentiate the inner function:*g*'(*x*).

## Proof

Start from *y* = *f*(*g*(*x*)).

Set *t* = *g*(*x*).

Then *dt*/*dx* = *g*'(*x*).

Put *t* = *g*(*x*) into *y* = *f*(*g*(*x*)).

Then *y* = *f*(*t*).

Then *dy*/*dt* = *f*'(*x*).

*y* = *f*(*g*(*x*))

So [*f*(*g*(*x*))]' = *dy*/*dx*

= (*dy*/*dt*)⋅(*dy*/*dx*).

Put *f*'(*t*) and *g*'(*x*)

into (*dy*/*dt*) and (*dt*/*dx*).

Put *g*(*x*) into the *t*.

Then [*f*(*g*(*x*))]' = *f*'(*g*(*x*))⋅*g*(*x*).

## Example 1

*y* = (2*x*^{2} - 1)^{8}*y*' is equal to,

the derivative of the outer part, 8(2*x*^{2} - 1)^{7}

times, the derivative of the inner part, (2⋅2*x*^{1} - 0).

Derivatives of polynomials

## Example 2

Previously, you've solved this example.

Let's solve the same example

by using the chain rule.

Change 1/(*x*^{3} + 2*x*) into (*x*^{3} + 2*x*)^{-1}.

*y* = (*x*^{3} + 2*x*)^{-1}*y*' is equal to,

the derivative of the outer part, (-1)⋅(*x*^{3} + 2*x*)^{-2}

times, the derivative of the inner part, (3⋅2*x*^{1} + 2⋅1*x*^{0}).

Derivatives of polynomials

As you can see,

you got the same answer.

So, to differentiate a function in reciprocal form,

you can either use the reciprocal rule

or use the chain rule.