Product Rule and Quotient Rule

Introduction

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

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

But, what about more complicated functions?

For example, what is  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)}   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

Calculate  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).}

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

Solution:
Using the Product 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 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
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)=(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
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{(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
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)=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:
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)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4)}
or equivalently
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)=x^5+3x^4+3x^3+6x^2+4x+4}

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 f(x)=\frac{x^2+x^3}{x}}

Solution:
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 f'(x)=\frac{x(x^2+x^3)'-(x^2+x^3)(x)'}{x^2}.}
Then, using the Power 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 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
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+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
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)=1+2x.}

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

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 f(x)=\frac{\sin x}{\cos x}}

Solution:
Using the Quotient Rule, 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{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  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 \sin^2 x+\cos^2 x=1}   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 \sec x=\frac{1}{\cos x}.}
Since  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{\sin x}{\cos x}=\tan x,}   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 \frac{d}{dx}{\tan x}=\sec^2 x.}

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 f'(x)=\sec^2 x}

Exercise 1

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{1}{x^2}(\csc x-4).}

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

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{\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  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} -substitution. 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=\frac{y}{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=\frac{1}{2}~dy}   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 2~du=dy.}

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{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

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=\arctan\bigg(\frac{y}{2}\bigg)+C.}

Exercise 2

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 g(x)=2x\sin x \sec 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=5+\sin(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 u=\cos(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{\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

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

Exercise 3

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

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,  $du=2~dx$  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

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