# Chain Rule

## Introduction

It is relatively easy to calculate the derivatives of simple functions, like polynomials or trigonometric functions.

But, what about more complicated functions?

For example,  ${\displaystyle f(x)=\sin(3x)}$  or  ${\displaystyle g(x)=(x+1)^{8}?}$

Well, the key to calculating the derivatives of these functions is to recognize that these functions are compositions.

For  ${\displaystyle f(x)=\sin(3x),}$  it is the composition of the function  ${\displaystyle y=3x}$  with  ${\displaystyle y=\sin(x).}$

Similarly, for  ${\displaystyle g(x)=(x+1)^{8},}$  it is the composition of  ${\displaystyle y=x+1}$  and  ${\displaystyle y=x^{8}.}$

So, how do we take the derivative of compositions?

The answer to this question is exactly the Chain Rule.

Chain Rule

Let  ${\displaystyle y=f(u)}$  be a differentiable function of  ${\displaystyle u}$  and let  ${\displaystyle u=g(x)}$  be a differentiable function of  ${\displaystyle x.}$

Then,  ${\displaystyle y=f\circ g(x))}$  is a differentiable function of  ${\displaystyle x}$  and

${\displaystyle y'=(f'\circ g(x))\cdot g'(x).}$

## Warm-Up

Calculate  ${\displaystyle h'(x).}$

1)   ${\displaystyle h(x)=\sin(3x)}$

Solution:
Let  ${\displaystyle f(x)=\sin(x)}$  and  ${\displaystyle g(x)=3x.}$
Then,  ${\displaystyle f'(x)=\cos(x)}$  and  ${\displaystyle g'(x)=3.}$
Now,  ${\displaystyle h(x)=f\circ g(x).}$
Using the Chain Rule, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {(f'\circ g(x))\cdot g'(x)}\\&&\\&=&\displaystyle {\cos(3x)\cdot 3}\\&&\\&=&\displaystyle {3\cos(3x).}\end{array}}}$
${\displaystyle h'(x)=3\cos(3x)}$

2)   ${\displaystyle h(x)=(x+1)^{8}}$

Solution:
Let  ${\displaystyle f(x)=x^{8}}$  and  ${\displaystyle g(x)=x+1.}$
Then,  ${\displaystyle f'(x)=8x^{7}}$  and  ${\displaystyle g'(x)=1.}$
Now,  ${\displaystyle h(x)=f\circ g(x).}$
Using the Chain Rule, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {(f'\circ g(x))\cdot g'(x)}\\&&\\&=&\displaystyle {8(x+1)^{7}\cdot 1}\\&&\\&=&\displaystyle {8(x+1)^{7}.}\end{array}}}$
${\displaystyle h'(x)=8(x+1)^{7}.}$

3)   ${\displaystyle h(x)=\ln(x^{2})}$

Solution:
Let  ${\displaystyle f(x)=\ln(x)}$  and  ${\displaystyle g(x)=x^{2}.}$
Then,  ${\displaystyle f'(x)={\frac {1}{x}}}$  and  ${\displaystyle g'(x)=2x.}$
Now,  ${\displaystyle h(x)=f\circ g(x).}$
Using the Chain Rule, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {(f'\circ g(x))\cdot g'(x)}\\&&\\&=&\displaystyle {{\frac {1}{x^{2}}}\cdot 2x}\\&&\\&=&\displaystyle {{\frac {2}{x}}.}\end{array}}}$
${\displaystyle h'(x)={\frac {2}{x}}}$

## Exercise 1

Calculate the derivative of  ${\displaystyle h(x)=(\sin x+\cos x)^{4}.}$

Using the Chain Rule, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {4(\sin x+\cos x)^{3}(\sin x+\cos x)'}\\&&\\&=&\displaystyle {4(\sin x+\cos x)^{3}((\sin x)'+(\cos x)')}\\&&\\&=&\displaystyle {4(\sin x+\cos x)^{3}(\cos x-\sin x)}.\end{array}}}$

So, we have

${\displaystyle h'(x)=4(\sin x+\cos x)^{3}(\cos x-\sin x).}$

## Exercise 2

Calculate the derivative of  ${\displaystyle h(x)=\sin ^{3}(2x^{2}+x+1).}$

First, notice  ${\displaystyle h(x)=(\sin(2x^{2}+x+1))^{3}.}$

Using the Chain Rule, we have

${\displaystyle h'(x)=3(\sin(2x^{2}+x+1))^{2}\cdot (\sin(2x^{2}+x+1))'.}$

Now, we need to use the Chain Rule a second time. So, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {3(\sin(2x^{2}+x+1))^{2}\cos(2x^{2}+x+1)\cdot (2x^{2}+x+1)'}\\&&\\&=&\displaystyle {3\sin ^{2}(2x^{2}+x+1)\cos(2x^{2}+x+1)(4x+1).}\end{array}}}$

So, we have

${\displaystyle h'(x)=3\sin ^{2}(2x^{2}+x+1)\cos(2x^{2}+x+1)(4x+1).}$

## Exercise 3

Calculate the derivative of  ${\displaystyle h(x)=\cos(2x+1)\sin(x^{2}+3x).}$

Using the Product Rule, we have

${\displaystyle h'(x)=\cos(2x+1)(\sin(x^{2}+3x))'+(\cos(2x+1))'\sin(x^{2}+3x).}$

For the two remaining derivatives, we need to use the Chain Rule.

So, using the Chain Rule, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\cos(2x+1)\cos(x^{2}+3x)\cdot (x^{2}+3x)'-\sin(2x+1)\cdot (2x+1)'\sin(x^{2}+3x)}\\&&\\&=&\displaystyle {\cos(2x+1)\cos(x^{2}+3x)(2x+3)-\sin(2x+1)(2)\sin(x^{2}+3x).}\end{array}}}$

So, we get

${\displaystyle h'(x)=\cos(2x+1)\cos(x^{2}+3x)(2x+3)-\sin(2x+1)(2)\sin(x^{2}+3x).}$

## Exercise 4

Calculate the derivative of  ${\displaystyle h(x)={\frac {\sin(3x)+x\cos(2x)}{x^{2}+1}}.}$

First, using the Quotient Rule, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\frac {(x^{2}+1)(\sin(3x)+x\cos(2x))'-(\sin(3x)+x\cos(2x))(x^{2}+1)'}{(x^{2}+1)^{2}}}\\&&\\&=&\displaystyle {{\frac {(x^{2}+1)[(\sin(3x))'+(x\cos(2x))']-(\sin(3x)+x\cos(2x))(2x)}{(x^{2}+1)^{2}}}.}\end{array}}}$

Using the Product Rule, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\frac {(x^{2}+1)[(\sin(3x))'+x(\cos(2x))'+(x)'\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^{2}+1)^{2}}}\\&&\\&=&\displaystyle {{\frac {(x^{2}+1)[(\sin(3x))'+x(\cos(2x))'+1\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^{2}+1)^{2}}}.}\end{array}}}$

For the remaining derivatives, we need to use the Chain Rule. So, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\frac {(x^{2}+1)[\cos(3x)(3x)'+x(-\sin(2x))(2x)'+\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^{2}+1)^{2}}}\\&&\\&=&\displaystyle {{\frac {(x^{2}+1)[\cos(3x)(3)-x\sin(2x)(2)+\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^{2}+1)^{2}}}.}\end{array}}}$

So, we have

${\displaystyle h'(x)={\frac {(x^{2}+1)[\cos(3x)(3)-x\sin(2x)(2)+\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^{2}+1)^{2}}}.}$