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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\sin(3x),}   it is the composition of the function  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=3x}   with  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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  $h(x)=(\sin x+\cos x)^{4}.$

Using the Chain Rule, we have

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

So, we have

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

Exercise 2

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

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

Using the Chain Rule, we have

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

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

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

So, we have

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

Exercise 3

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

Using the Product Rule, we have

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

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

So, using the Chain Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\frac{e^x}{x^2\sin x}.}

First, using the Quotient Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 x e^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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.} \end{array}}

So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}