Product Rule and Quotient Rule - Revision history

MathAdmin: Created page with "==Introduction== Taking the derivatives of simple functions (i.e. polynomials) is easy using the power rule. For example, if
2017-10-05T21:59:44Z

Created page with "==Introduction== Taking the derivatives of <em>simple functions</em> (i.e. polynomials) is easy using the power rule. For example, if <math style="vertical-align: -5px..."

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==Introduction==
Taking the derivatives of <em>simple functions</em> (i.e. polynomials) is easy using the power rule.

For example, if  <math style="vertical-align: -5px">f(x)=x^3+2x^2+5x+3,</math>  then  <math style="vertical-align: -5px">f'(x)=3x^2+4x+5.</math>

But, what about more <em>complicated functions</em>?

For example, what is  <math style="vertical-align: -5px">f'(x)</math>  when  <math style="vertical-align: -5px">f(x)=\sin x \cos x?</math>

Or what about  <math style="vertical-align: -5px">g'(x)</math>  when  <math style="vertical-align: -15px">g(x)=\frac{x}{x+1}?</math>

Notice  <math style="vertical-align: -5px">f(x)</math>  is a product, and  <math style="vertical-align: -5px">g(x)</math>  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  <math style="vertical-align: -5px">h(x)=f(x)g(x).</math>  Then,

::<math>h'(x)=f(x)g'(x)+f'(x)g(x).</math>

'''Quotient Rule'''

Let  <math style="vertical-align: -19px">h(x)=\frac{f(x)}{g(x)}.</math>  Then,

::<math>h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}.</math>

==Warm-Up==
Calculate  <math style="vertical-align: -5px">f'(x).</math>

'''1)'''   <math style="vertical-align: -7px">f(x)=(x^2+x+1)(x^3+2x^2+4)</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Using the Product Rule, we have
|-
|
::<math>f'(x)=(x^2+x+1)(x^3+2x^2+4)'+(x^2+x+1)'(x^3+2x^2+4).</math>
|-
|Then, using the Power Rule, we have
|-
|
::<math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4).</math>
|-
|-
|<u>NOTE:</u> 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
|-
|
::<math>\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}</math>
|-
|Now, using the Power Rule, we get
|-
|
::<math>f'(x)=5x^4+12x^3+9x^2+12x+4.</math>
|-
|In general, calculating derivatives in this way is tedious. It would be better to use the Product Rule.
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4)</math>
|-
|or equivalently
|-
|       <math>f'(x)=x^5+3x^4+3x^3+6x^2+4x+4</math>
|}

'''2)'''   <math style="vertical-align: -14px">f(x)=\frac{x^2+x^3}{x}</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|
Using the Quotient Rule, we have
|-
|
::<math>f'(x)=\frac{x(x^2+x^3)'-(x^2+x^3)(x)'}{x^2}.</math>
|-
|Then, using the Power Rule, we have
|-
|
::<math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)(1)}{x^2}.</math>
|-
|<u>NOTE:</u> 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
|-
|
::<math>\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}</math>
|-
|Now, using the Power Rule, we get
|-
|
::<math>f'(x)=1+2x.</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
||       <math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)}{x^2}</math>
|-
|or equivalently
|-
|       <math>f'(x)=1+2x</math>

|-
|}

'''3)'''   <math style="vertical-align: -14px">f(x)=\frac{\sin x}{\cos x}</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Using the Quotient Rule, we get
|-
|
::<math>\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}</math>
|-
|since  <math style="vertical-align: -2px">\sin^2 x+\cos^2 x=1</math>  and  <math style="vertical-align: -13px">\sec x=\frac{1}{\cos x}.</math>
|-
|Since  <math style="vertical-align: -14px">\frac{\sin x}{\cos x}=\tan x,</math>  we have
|-
|
::<math>\frac{d}{dx}{\tan x}=\sec^2 x.</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>f'(x)=\sec^2 x</math>
|-
|}

== Exercise 1 ==

Calculate the derivative of  <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}(\csc x-4).</math>

First, we need to know the derivative of  <math style="vertical-align: 0px">\csc x.</math>  Recall

::<math>\csc x =\frac{1}{\sin x}.</math>

Now, using the Quotient Rule, we have

::<math>\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}</math>

Using the Product Rule and Power Rule, we have

::<math>\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}</math>

So, we have
::<math>f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.</math>

== Exercise 2 ==

Calculate the derivative of  <math style="vertical-align: -5px">g(x)=2x\sin x \sec x.</math>

Notice that the function  <math style="vertical-align: -5px">g(x)</math>  is the product of three functions.

We start by grouping two of the functions together. So, we have  <math style="vertical-align: -5px">g(x)=(2x\sin x)\sec x.</math>

Using the Product Rule, we get

::<math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\
&&\\
& = & \displaystyle{(2x\sin x)(\tan^2 x)+(2x\sin x)'\sec x.}
\end{array}</math>

Now, we need to use the Product Rule again. So,

::<math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{2x\sin x\tan^2 x+(2x(\sin x)'+(2x)'\sin x)\sec x}\\
&&\\
& = & \displaystyle{2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}
\end{array}</math>

So, we have
::<math>g'(x)=2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.</math>

But, there is another way to do this problem. Notice

::<math>\begin{array}{rcl}
\displaystyle{g(x)} & = & \displaystyle{2x\sin x\sec x}\\
&&\\
& = & \displaystyle{2x\sin x\frac{1}{\cos x}}\\
&&\\
& = & \displaystyle{2x\tan x.}
\end{array}</math>

Now, you would only need to use the Product Rule once instead of twice.

== Exercise 3 ==

Calculate the derivative of  <math style="vertical-align: -16px">h(x)=\frac{x^2\sin x+1}{x^2\cos x+3}.</math>

Using the Quotient Rule, we have

::<math>h'(x)=\frac{(x^2\cos x+3)(x^2\sin x+1)'-(x^2\sin x+1)(x^2\cos x+3)'}{(x^2\cos x+3)^2}.</math>

Now, we need to use the Product Rule. So, we have

::<math>\begin{array}{rcl}
\displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2\cos x+3)(x^2(\sin x)'+(x^2)'\sin x)-(x^2\sin x+1)(x^2(\cos x)'+(x^2)'\cos x)}{(x^2\cos x+3)^2}}\\
&&\\
& = & \displaystyle{\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.}
\end{array}</math>

So, we get
::<math>h'(x)=\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.</math>

== Exercise 4 ==

Calculate the derivative of  <math style="vertical-align: -14px">f(x)=\frac{e^x}{x^2\sin x}.</math>

First, using the Quotient Rule, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
\end{array}</math>

Now, we need to use the Product Rule. So, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
\end{array}</math>

So, we have
::<math>f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.</math>