Product Rule and Quotient Rule

Introduction

Taking the derivatives of simple functions (i.e. polynomials) is easy using the power rule.

For example, if  $f(x)=x^{3}+2x^{2}+5x+3,$  then  $f'(x)=3x^{2}+4x+5.$

But, what about more complicated functions?

For example, what is  $f'(x)$  when  $f(x)=\sin x\cos x?$

Or what about  $g'(x)$  when  $g(x)={\frac {x}{x+1}}?$

Notice  $f(x)$  is a product and  $g(x)$  is a quotient. So, to answer the question of how to calculate these derivatives, we look to the Product Rule and the Quotient Rule. The Product Rule and the Quotient Rule give us formulas for calculating these derivatives.

Product Rule

Let  $h(x)=f(x)g(x).$  Then,

$h'(x)=f(x)g'(x)+f'(x)g(x).$

Quotient Rule

Let  $h(x)={\frac {f(x)}{g(x)}}.$  Then,

$h'(x)={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}.$

Warm-Up

Calculate  $f'(x).$

1)   $f(x)=(x^{2}+x+1)(x^{3}+2x^{2}+4)$

Solution:
Using the Product Rule, we have
$f'(x)=(x^{2}+x+1)(x^{3}+2x^{2}+4)'+(x^{2}+x+1)'(x^{3}+2x^{2}+4).$
Then, using the Power Rule, we have
$f'(x)=(x^{2}+x+1)(3x^{2}+4x)+(2x+1)(x^{3}+2x^{2}+4).$
NOTE: It is not necessary to use the Product Rule to calculate the derivative of this function.
You can distribute the terms and then use the Power Rule.
In this case, we have
${\begin{array}{rcl}\displaystyle {f(x)}&=&\displaystyle {(x^{2}+x+1)(x^{3}+2x^{2}+4)}\\&&\\&=&\displaystyle {x^{2}(x^{3}+2x^{2}+4)+x(x^{3}+2x^{2}+4)+1(x^{3}+2x^{2}+4)}\\&&\\&=&\displaystyle {x^{5}+2x^{4}+4x^{2}+x^{4}+2x^{3}+4x+x^{3}+2x^{2}+4}\\&&\\&=&\displaystyle {x^{5}+3x^{4}+3x^{3}+6x^{2}+4x+4.}\end{array}}$
Now, using the Power Rule, we get
$f'(x)=5x^{4}+12x^{3}+9x^{2}+12x+4.$
In general, calculating derivatives in this way is tedious. It would be better to use the Product Rule.

Final Answer:
$f'(x)=(x^{2}+x+1)(3x^{2}+4x)+(2x+1)(x^{3}+2x^{2}+4)$
or equivalently
$f'(x)=x^{5}+3x^{4}+3x^{3}+6x^{2}+4x+4$

2)   $f(x)={\frac {x^{2}+x^{3}}{x}}$

Solution:
Using the Quotient Rule, we have
$f'(x)={\frac {x(x^{2}+x^{3})'-(x^{2}+x^{3})(x)'}{x^{2}}}.$
Then, using the Power Rule, we have
$f'(x)={\frac {x(2x+3x^{2})-(x^{2}+x^{3})(1)}{x^{2}}}.$
NOTE: It is not necessary to use the Quotient Rule to calculate the derivative of this function.
You can divide and then use the Power Rule.
In this case, we have
${\begin{array}{rcl}\displaystyle {f(x)}&=&\displaystyle {\frac {x^{2}+x^{3}}{x}}\\&&\\&=&\displaystyle {{\frac {x^{2}}{x}}+{\frac {x^{3}}{x}}}\\&&\\&=&\displaystyle {x+x^{2}.}\\\end{array}}$
Now, using the Power Rule, we get
$f'(x)=1+2x.$

Final Answer:
$f'(x)={\frac {x(2x+3x^{2})-(x^{2}+x^{3})}{x^{2}}}$
or equivalently
$f'(x)=1+2x$

3)   $f(x)={\frac {\sin x}{\cos x}}$

Solution:
Using the Quotient Rule, we get
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {\cos x(\sin x)'-\sin x(\cos x)'}{(\cos x)^{2}}}\\&&\\&=&\displaystyle {\frac {\cos x(\cos x)-\sin x(-\sin x)}{(\cos x)^{2}}}\\&&\\&=&\displaystyle {\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&&\\&=&\displaystyle {\frac {1}{\cos ^{2}x}}\\&&\\&=&\displaystyle {\sec ^{2}x}\end{array}}$
since  $\sin ^{2}x+\cos ^{2}x=1$  and  $\sec x={\frac {1}{\cos x}}.$
Since  ${\frac {\sin x}{\cos x}}=\tan x,$  we have
${\frac {d}{dx}}{\tan x}=\sec ^{2}x.$

Final Answer:
$f'(x)=\sec ^{2}x$

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

$f'(x)={\frac {-\csc x\cot x}{x^{2}}}+{\frac {-2(\csc x-4)}{x^{3}}}.$

Exercise 2

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

Let  $u=5+\sin(x).$  Then,  $u=\cos(x)~dx.$

Plugging these into our integral, we get

${\begin{array}{rcl}\displaystyle {\int {\frac {\cos(x)}{(5+\sin x)^{2}}}~dx}&=&\displaystyle {\int {\frac {1}{u^{2}}}~du}\\&&\\&=&\displaystyle {-{\frac {1}{u}}+C}\\&&\\&=&\displaystyle {-{\frac {1}{5+\sin(x)}}+C.}\end{array}}$

So, we have

$\int {\frac {\cos(x)}{(5+\sin x)^{2}}}~dx=-{\frac {1}{5+\sin(x)}}+C.$

Exercise 3

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

Here, the substitution is not obvious.

Let  $u=2x+3.$  Then,  $du=2~dx$  and  ${\frac {du}{2}}=dx.$

Now, we need a way of getting rid of  $x+5$  in the numerator.

Solving for  $x$  in the first equation, 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 x=\frac{1}{2}u-\frac{3}{2}.}

Plugging these into our integral, 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 \begin{array}{rcl} \displaystyle{\int \frac{x+5}{2x+3}~dx} & = & \displaystyle{\int \frac{(\frac{1}{2}u-\frac{3}{2})+5}{2u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{2}\int \frac{\frac{1}{2}u+\frac{7}{2}}{u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{4}\int \frac{u+7}{u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{4}\int 1+\frac{7}{u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{4}(u+7\ln|u|)+C}\\ &&\\ & = & \displaystyle{\frac{1}{4}(2x+3+7\ln|2x+3|)+C.}\\ \end{array}}

So, we get

$\int {\frac {x+5}{2x+3}}~dx={\frac {1}{4}}(2x+3+7\ln |2x+3|)+C.$

Exercise 4

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

Let  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 u=x+2.}   Then,  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 du=dx.}

Now, we need a way of replacing  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 x^2+4.}

If we solve 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 x}   in our first equation, 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 x=u-2.}

Now, we square both sides of this last equation to 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 x^2=(u-2)^2.}

Plugging in to our integral, 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 \begin{array}{rcl} \displaystyle{\int \frac{x^2+4}{x+2}~dx} & = & \displaystyle{\int \frac{(u-2)^2+4}{u}~du}\\ &&\\ & = & \displaystyle{\int \frac{u^2-4u+4+4}{u}~du}\\ &&\\ & = & \displaystyle{\int \frac{u^2-4u+8}{u}~du}\\ &&\\ & = & \displaystyle{\int u-4+\frac{8}{u}~du}\\ &&\\ & = & \displaystyle{\frac{u^2}{2}-4u+8\ln|u|+C}\\ &&\\ & = & \displaystyle{\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.}\\ \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 \int \frac{x^2+4}{x+2}~dx=\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.}