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  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 x \cos x?}

Or what about  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)}   when  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)=\frac{x}{x+1}?}

Notice  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)}   is a product and  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)}   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  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)=f(x)g(x).}   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 h'(x)=f(x)g'(x)+f'(x)g(x).}

Quotient Rule

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

Warm-Up

Evaluate the following indefinite integrals.

1)   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 (8x+5)e^{4x^2+5x+3}~dx}

Solution:
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=4x^2+5x+3.}   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=(8x+5)~dx.}
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 \int e^u~du,}   which we know how to integrate.
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 \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:
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 e^{4x^2+5x+3}+C}

2)   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}{\sqrt{1-2x^2}}~dx}

Solution:
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=1-2x^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=-4x~dx.}   Hence,  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 \frac{du}{-4}=x~dx.}
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}{\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:
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 -\frac{1}{2}\sqrt{1-2x^2}+C}

3)   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{\sin(\ln x)}{x}~dx}

Solution:
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=\ln(x).}   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=\frac{1}{x}~dx.}
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{\sin(\ln x)}{x}~dx} & = & \displaystyle{\int \sin(u)~du}\\ &&\\ & = & \displaystyle{-\cos(u)+C}\\ &&\\ & = & \displaystyle{-\cos(\ln x)+C.} \\ \end{array}}

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

4)   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 xe^{x^2}~dx}

Solution:
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=2x~dx}   and  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 \frac{du}{2}=x~dx.}
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 xe^{x^2}~dx} & = & \displaystyle{\int \frac{1}{2}e^u~du}\\ &&\\ & = & \displaystyle{\frac{1}{2}e^u+C}\\ &&\\ & = & \displaystyle{\frac{1}{2}e^{x^2}+C.} \\ \end{array}}

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

Exercise 1

Evaluate the indefinite integral  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{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  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=2x+3.}   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=2~dx}   and  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 \frac{du}{2}=dx.}

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

Solving 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 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

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

Exercise 4

Evaluate the indefinite integral  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.}

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.}