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

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

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

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

So, how do we take the derivative of compositions?

The answer to this question is exactly the Chain Rule.

Chain Rule

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

Then,  $y=f(g(x))$  is a differentiable function of  $x$  and

$y'=f'(g(x))\cdot g'(x).$

Warm-Up

Calculate  $h'(x).$

1)   $h(x)=\sin(3x)$

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

Final Answer:
$h'(x)=3\cos(3x)$

2)   $h(x)=(x+1)^{8}$

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

Final Answer:
$h'(x)=8(x+1)^{7}.$

3)   $h(x)=\ln(x^{2})$

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

Final Answer:
$h'(x)={\frac {2}{x}}$

Exercise 1

Calculate the derivative of  $f(x)={\frac {1}{x^{2}}}(\csc x-4).$

First, we need to know the derivative of  $\csc x.$  Recall

$\csc x={\frac {1}{\sin x}}.$

Now, using the Quotient Rule, we have

${\begin{array}{rcl}\displaystyle {{\frac {d}{dx}}(\csc x)}&=&\displaystyle {{\frac {d}{dx}}{\bigg (}{\frac {1}{\sin x}}{\bigg )}}\\&&\\&=&\displaystyle {\frac {\sin x(1)'-1(\sin x)'}{\sin ^{2}x}}\\&&\\&=&\displaystyle {\frac {\sin x(0)-\cos x}{\sin ^{2}x}}\\&&\\&=&\displaystyle {\frac {-\cos x}{\sin ^{2}x}}\\&&\\&=&\displaystyle {-\csc x\cot x.}\end{array}}$

Using the Product Rule and Power Rule, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {1}{x^{2}}}(\csc x-4)'+{\bigg (}{\frac {1}{x^{2}}}{\bigg )}'(\csc x-4)}\\&&\\&=&\displaystyle {{\frac {1}{x^{2}}}(-\csc x\cot x+0)+(-2x^{-3})(\csc x-4)}\\&&\\&=&\displaystyle {{\frac {-\csc x\cot x}{x^{2}}}+{\frac {-2(\csc x-4)}{x^{3}}}.}\end{array}}$

So, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)={\frac {-\csc x\cot x}{x^{2}}}+{\frac {-2(\csc x-4)}{x^{3}}}.}

Exercise 2

Calculate the derivative of  $g(x)=2x\sin x\sec x.$

Notice that the function  $g(x)$  is the product of three functions.

We start by grouping two of the functions together. So, we have  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=(2x\sin x)\sec x.}

Using the Product Rule, we get

${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\&&\\&=&\displaystyle {(2x\sin x)(\tan ^{2}x)+(2x\sin x)'\sec x.}\end{array}}$

Now, we need to use the Product Rule again. So,

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {2x\sin x\tan ^{2}x+(2x(\sin x)'+(2x)'\sin x)\sec x}\\&&\\&=&\displaystyle {2x\sin x\tan ^{2}x+(2x\cos x+2\sin x)\sec x.}\end{array}}}

So, we have

$g'(x)=2x\sin x\tan ^{2}x+(2x\cos x+2\sin x)\sec x.$

But, there is another way to do this problem. Notice

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {g(x)}&=&\displaystyle {2x\sin x\sec x}\\&&\\&=&\displaystyle {2x\sin x{\frac {1}{\cos x}}}\\&&\\&=&\displaystyle {2x\tan x.}\end{array}}}

Now, you would only need to use the Product Rule once instead of twice.

Exercise 3

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

Using the Quotient Rule, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h'(x)={\frac {(x^{2}\cos x+3)(x^{2}\sin x+1)'-(x^{2}\sin x+1)(x^{2}\cos x+3)'}{(x^{2}\cos x+3)^{2}}}.}

Now, we need to use the Product Rule. So, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\frac {(x^{2}\cos x+3)(x^{2}(\sin x)'+(x^{2})'\sin x)-(x^{2}\sin x+1)(x^{2}(\cos x)'+(x^{2})'\cos x)}{(x^{2}\cos x+3)^{2}}}\\&&\\&=&\displaystyle {{\frac {(x^{2}\cos x+3)(x^{2}\cos x+2x\sin x)-(x^{2}\sin x+1)(-x^{2}\sin x+2x\cos x)}{(x^{2}\cos x+3)^{2}}}.}\end{array}}}

So, we get

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

Exercise 4

Calculate the derivative of  $f(x)={\frac {e^{x}}{x^{2}\sin x}}.$

First, using the Quotient Rule, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin x(e^{x})'-e^{x}(x^{2}\sin x)'}{(x^{2}\sin x)^{2}}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\sin x)'}{x^{4}\sin ^{2}x}}.}\end{array}}$

Now, we need to use the Product Rule. So, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}$

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$