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:
Let  $u=4x^{2}+5x+3.$  Then,  $du=(8x+5)~dx.$
Plugging these into our integral, we get  $\int e^{u}~du,$  which we know how to integrate.
So, we get
${\begin{array}{rcl}\displaystyle {\int (8x+5)e^{4x^{2}+5x+3}~dx}&=&\displaystyle {\int e^{u}~du}\\&&\\&=&\displaystyle {e^{u}+C}\\&&\\&=&\displaystyle {e^{4x^{2}+5x+3}+C.}\\\end{array}}$

Final Answer:
$e^{4x^{2}+5x+3}+C$

2)   $f(x)={\frac {x^{2}+x^{3}}{x}}$

Solution:
Let  $u=1-2x^{2}.$  Then,  $du=-4x~dx.$  Hence,  ${\frac {du}{-4}}=x~dx.$
Plugging these into our integral, we get
${\begin{array}{rcl}\displaystyle {\int {\frac {x}{\sqrt {1-2x^{2}}}}~dx}&=&\displaystyle {\int -{\frac {1}{4}}~u^{-{\frac {1}{2}}}~du}\\&&\\&=&\displaystyle {-{\frac {1}{2}}u^{\frac {1}{2}}+C}\\&&\\&=&\displaystyle {-{\frac {1}{2}}{\sqrt {1-2x^{2}}}+C.}\\\end{array}}$

Final Answer:
$-{\frac {1}{2}}{\sqrt {1-2x^{2}}}+C$

3)   $f(x)={\frac {\sin x}{\cos x}}$

Solution:
Let  $u=\ln(x).$  Then,  $du={\frac {1}{x}}~dx.$
Plugging these into our integral, we get
${\begin{array}{rcl}\displaystyle {\int {\frac {\sin(\ln x)}{x}}~dx}&=&\displaystyle {\int \sin(u)~du}\\&&\\&=&\displaystyle {-\cos(u)+C}\\&&\\&=&\displaystyle {-\cos(\ln x)+C.}\\\end{array}}$

Final Answer:
$-\cos(\ln x)+C$

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

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