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

Evaluate the indefinite integral  $\int {\frac {2}{y^{2}+4}}~dy.$

First, we factor out  $4$  out of the denominator.

So, we have

${\begin{array}{rcl}\displaystyle {\int {\frac {2}{y^{2}+4}}~dy}&=&\displaystyle {{\frac {1}{4}}\int {\frac {2}{{\frac {y^{2}}{4}}+1}}~dy}\\&&\\&=&\displaystyle {{\frac {1}{2}}\int {\frac {1}{({\frac {y}{2}})^{2}+1}}~dy.}\\\end{array}}$

Now, we use  $u$-substitution. Let  $u={\frac {y}{2}}.$

Then,  $du={\frac {1}{2}}~dy$  and  $2~du=dy.$

Plugging these into our integral, we get

${\begin{array}{rcl}\displaystyle {\int {\frac {2}{y^{2}+4}}~dy}&=&\displaystyle {{\frac {1}{2}}\int {\frac {2}{u^{2}+1}}~du}\\&&\\&=&\displaystyle {\int {\frac {1}{u^{2}+1}}~du}\\&&\\&=&\displaystyle {\arctan(u)+C}\\&&\\&=&\displaystyle {\arctan {\bigg (}{\frac {y}{2}}{\bigg )}+C.}\\\end{array}}$

So, we have

$\int {\frac {2}{y^{2}+4}}~dy=\arctan {\bigg (}{\frac {y}{2}}{\bigg )}+C.$

Exercise 2

Evaluate the indefinite integral  $\int {\frac {\cos(x)}{(5+\sin x)^{2}}}~dx.$

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

Evaluate the indefinite integral  $\int {\frac {x+5}{2x+3}}~dx.$

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  $x={\frac {1}{2}}u-{\frac {3}{2}}.$

Plugging these into our integral, we get

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

So, we get

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

Exercise 4

Evaluate the indefinite integral  $\int {\frac {x^{2}+4}{x+2}}~dx.$

Let  $u=x+2.$  Then,  $du=dx.$

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

If we solve for  $x$  in our first equation, we get  $x=u-2.$

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

Plugging in to our integral, we get

${\begin{array}{rcl}\displaystyle {\int {\frac {x^{2}+4}{x+2}}~dx}&=&\displaystyle {\int {\frac {(u-2)^{2}+4}{u}}~du}\\&&\\&=&\displaystyle {\int {\frac {u^{2}-4u+4+4}{u}}~du}\\&&\\&=&\displaystyle {\int {\frac {u^{2}-4u+8}{u}}~du}\\&&\\&=&\displaystyle {\int u-4+{\frac {8}{u}}~du}\\&&\\&=&\displaystyle {{\frac {u^{2}}{2}}-4u+8\ln |u|+C}\\&&\\&=&\displaystyle {{\frac {(x+2)^{2}}{2}}-4(x+2)+8\ln |x+2|+C.}\\\end{array}}$

So, we have

$\int {\frac {x^{2}+4}{x+2}}~dx={\frac {(x+2)^{2}}{2}}-4(x+2)+8\ln |x+2|+C.$