Difference between revisions of "Product Rule and Quotient Rule"

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Taking the derivatives of <em>simple functions</em> (i.e. polynomials) is easy using the power rule.  
 
Taking the derivatives of <em>simple functions</em> (i.e. polynomials) is easy using the power rule.  
  
−
For example, if &nbsp;<math>f(x)=x^3+2x^2+5x+3,</math>&nbsp; then &nbsp;<math>f'(x)=3x^2+4x+5.</math>
+
For example, if &nbsp;<math style="vertical-align: -5px">f(x)=x^3+2x^2+5x+3,</math>&nbsp; then &nbsp;<math style="vertical-align: -5px">f'(x)=3x^2+4x+5.</math>
  
 
But, what about more <em>complicated functions</em>?  
 
But, what about more <em>complicated functions</em>?  
  
−
For example, what is &nbsp;<math>f'(x)</math>&nbsp; when &nbsp;<math>f(x)=\sin x \cos x?</math>
+
For example, what is &nbsp;<math style="vertical-align: -5px">f'(x)</math>&nbsp; when &nbsp;<math style="vertical-align: -5px">f(x)=\sin x \cos x?</math>
  
−
Or what about &nbsp;<math>g'(x)</math>&nbsp; when &nbsp;<math>g(x)=\frac{x}{x+1}?</math>
+
Or what about &nbsp;<math style="vertical-align: -5px">g'(x)</math>&nbsp; when &nbsp;<math style="vertical-align: -15px">g(x)=\frac{x}{x+1}?</math>
  
−
Notice &nbsp;<math>f(x)</math>&nbsp; is a product and &nbsp;<math>g(x)</math>&nbsp; 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.
+
Notice &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is a product, and &nbsp;<math style="vertical-align: -5px">g(x)</math>&nbsp; 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'''
 
'''Product Rule'''
  
−
Let &nbsp;<math>h(x)=f(x)g(x).</math>&nbsp; Then,
+
Let &nbsp;<math style="vertical-align: -5px">h(x)=f(x)g(x).</math>&nbsp; Then,
  
 
::<math>h'(x)=f(x)g'(x)+f'(x)g(x).</math>
 
::<math>h'(x)=f(x)g'(x)+f'(x)g(x).</math>
Line 20: Line 20:
 
'''Quotient Rule'''
 
'''Quotient Rule'''
  
−
Let &nbsp;<math>h(x)=\frac{f(x)}{g(x)}.</math>&nbsp; Then,
+
Let &nbsp;<math style="vertical-align: -19px">h(x)=\frac{f(x)}{g(x)}.</math>&nbsp; Then,
  
 
::<math>h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}.</math>
 
::<math>h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}.</math>
  
 
==Warm-Up==
 
==Warm-Up==
−
Evaluate the following indefinite integrals.
+
Calculate &nbsp;<math style="vertical-align: -5px">f'(x).</math>
  
−
'''1)''' &nbsp; <math>\int (8x+5)e^{4x^2+5x+3}~dx</math>
+
'''1)''' &nbsp; <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;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Let &nbsp;<math style="vertical-align: -3px">u=4x^2+5x+3.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">du=(8x+5)~dx.</math>
+
|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>
 
|-
 
|-
−
|Plugging these into our integral, we get &nbsp;<math style="vertical-align: -14px">\int e^u~du,</math>&nbsp; which we know how to integrate.
+
|Then, using the Power Rule, we have
 
|-
 
|-
−
|So, we get
+
|
 +
::<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}
 
::<math>\begin{array}{rcl}
−
\displaystyle{\int (8x+5)e^{4x^2+5x+3}~dx} & = & \displaystyle{\int e^u~du}\\
+
\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{e^u+C}\\
+
& = & \displaystyle{x^5+2x^4+4x^2+x^4+2x^3+4x+x^3+2x^2+4} \\
 
&&\\
 
&&\\
−
& = & \displaystyle{e^{4x^2+5x+3}+C.} \\
+
& = & \displaystyle{x^5+3x^4+3x^3+6x^2+4x+4.}
 
\end{array}</math>
 
\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.
 
|}
 
|}
  
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>e^{4x^2+5x+3}+C</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4)</math>
 
|-
 
|-
 +
|or equivalently
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=x^5+3x^4+3x^3+6x^2+4x+4</math>
 
|}
 
|}
  
−
'''2)''' &nbsp; <math>\int\frac{x}{\sqrt{1-2x^2}}~dx</math>
+
'''2)''' &nbsp; <math style="vertical-align: -14px">f(x)=\frac{x^2+x^3}{x}</math>
  
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Let &nbsp;<math style="vertical-align: -2px">u=1-2x^2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -2px">du=-4x~dx.</math>&nbsp; Hence, &nbsp;<math style="vertical-align: -15px">\frac{du}{-4}=x~dx.</math>&nbsp;
+
|
−
|-
+
Using the Quotient Rule, we have
−
|Plugging these into our integral, we get
 
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
::<math>f'(x)=\frac{x(x^2+x^3)'-(x^2+x^3)(x)'}{x^2}.</math>
−
\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}</math>
 
 
|-
 
|-
−
|}
+
|Then, using the Power Rule, we have
−
 
 
−
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Final Answer: &nbsp;
 
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>-\frac{1}{2}\sqrt{1-2x^2}+C</math>
+
|
 +
::<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.
−
 
 
−
'''3)''' &nbsp; <math>\int\frac{\sin(\ln x)}{x}~dx</math>
 
−
 
 
−
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Solution: &nbsp;
 
 
|-
 
|-
−
|Let &nbsp;<math style="vertical-align: -6px">u=\ln(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -14px">du=\frac{1}{x}~dx.</math>
+
|You can divide and then use the Power Rule.
 
|-
 
|-
−
|Plugging these into our integral, we get
+
|In this case, we have
 
|-
 
|-
 
|
 
|
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{\int\frac{\sin(\ln x)}{x}~dx} & = & \displaystyle{\int \sin(u)~du}\\
+
\displaystyle{f(x)} & = & \displaystyle{\frac{x^2+x^3}{x}}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{-\cos(u)+C}\\
+
& = & \displaystyle{\frac{x^2}{x}+\frac{x^3}{x}}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{-\cos(\ln x)+C.} \\
+
& = & \displaystyle{x+x^2.} \\
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
 +
|Now, using the Power Rule, we get
 +
|-
 +
|
 +
::<math>f'(x)=1+2x.</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>-\cos(\ln x)+C</math>
+
||&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)}{x^2}</math>
 +
|-
 +
|or equivalently
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=1+2x</math>
 +
 
 
|-
 
|-
 
|}
 
|}
  
−
'''4)''' &nbsp; <math>\int xe^{x^2}~dx</math>
+
'''3)''' &nbsp; <math style="vertical-align: -14px">f(x)=\frac{\sin x}{\cos x}</math>
  
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|Let &nbsp;<math style="vertical-align: -1px">u=x^2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=2x~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -15px">\frac{du}{2}=x~dx.</math>&nbsp;
+
|Using the Quotient Rule, we get
−
|-
 
−
|Plugging these into our integral, we get
 
 
|-
 
|-
 
|
 
|
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{\int xe^{x^2}~dx} & = & \displaystyle{\int \frac{1}{2}e^u~du}\\
+
\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{1}{2}e^u+C}\\
+
& = & \displaystyle{\frac{\cos^2 x+\sin^2 x}{\cos^2 x}} \\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{1}{2}e^{x^2}+C.} \\
+
& = & \displaystyle{\frac{1}{\cos^2 x}}\\
 +
&&\\
 +
& = & \displaystyle{\sec^2 x}
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
 +
|since &nbsp;<math style="vertical-align: -2px">\sin^2 x+\cos^2 x=1</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\sec x=\frac{1}{\cos x}.</math>
 +
|-
 +
|Since &nbsp;<math style="vertical-align: -14px">\frac{\sin x}{\cos x}=\tan x,</math>&nbsp; we have
 +
|-
 +
|
 +
::<math>\frac{d}{dx}{\tan x}=\sec^2 x.</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{2}e^{x^2}+C</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\sec^2 x</math>
 
|-
 
|-
 
|}
 
|}
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== Exercise 1 ==
 
== Exercise 1 ==
  
−
Evaluate the indefinite integral &nbsp;<math style="vertical-align: -16px">\int \frac{2}{y^2+4}~dy.</math>
+
Calculate the derivative of &nbsp;<math style="vertical-align: -13px">f(x)=\frac{1}{x^2}(\csc x-4).</math>
  
−
First, we factor out &nbsp;<math style="vertical-align: -1px">4</math>&nbsp; out of the denominator.
+
First, we need to know the derivative of &nbsp;<math style="vertical-align: 0px">\csc x.</math>&nbsp; Recall
  
−
So, we have
+
::<math>\csc x =\frac{1}{\sin x}.</math>
 +
 
 +
Now, using the Quotient Rule, we have
  
 
::<math>\begin{array}{rcl}
 
::<math>\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{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{1}{2}\int \frac{1}{(\frac{y}{2})^2+1}~dy.}\\
+
& = & \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>
 
\end{array}</math>
  
−
Now, we use &nbsp;<math style="vertical-align: -1px">u</math>-substitution. Let &nbsp;<math>u=\frac{y}{2}.</math>
+
Using the Product Rule and Power Rule, we have
−
 
 
−
Then, &nbsp;<math style="vertical-align: -14px">du=\frac{1}{2}~dy</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">2~du=dy.</math>
 
−
 
 
−
Plugging these into our integral, we get
 
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{\int \frac{2}{y^2+4}~dy} & = & \displaystyle{\frac{1}{2}\int \frac{2}{u^2+1}~du}\\
+
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\
−
&&\\
 
−
& = & \displaystyle{\int \frac{1}{u^2+1}~du}\\
 
 
&&\\
 
&&\\
−
& = & \displaystyle{\arctan(u)+C}\\
+
& = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\arctan\bigg(\frac{y}{2}\bigg)+C.}\\
+
& = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}
 
\end{array}</math>
 
\end{array}</math>
  
 
So, we have  
 
So, we have  
−
::<math>\int \frac{2}{y^2+4}~dy=\arctan\bigg(\frac{y}{2}\bigg)+C.</math>
+
::<math>f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.</math>
  
 
== Exercise 2 ==
 
== Exercise 2 ==
  
−
Evaluate the indefinite integral &nbsp;<math style="vertical-align: -17px">\int \frac{\cos(x)}{(5+\sin x)^2}~dx.</math>
+
Calculate the derivative of &nbsp;<math style="vertical-align: -5px">g(x)=2x\sin x \sec x.</math>
  
−
Let &nbsp;<math style="vertical-align: -5px">u=5+\sin(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">u=\cos(x)~dx.</math>
+
Notice that the function &nbsp;<math style="vertical-align: -5px">g(x)</math>&nbsp; is the product of three functions.  
  
−
Plugging these into our integral, we get
+
We start by grouping two of the functions together. So, we have &nbsp;<math style="vertical-align: -5px">g(x)=(2x\sin x)\sec x.</math>
 +
 
 +
Using the Product Rule, we get
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{\int \frac{\cos(x)}{(5+\sin x)^2}~dx} & = & \displaystyle{\int \frac{1}{u^2}~du}\\
+
\displaystyle{g'(x)} & = & \displaystyle{(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{-\frac{1}{u}+C}\\
+
& = & \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{-\frac{1}{5+\sin(x)}+C.}
+
& = & \displaystyle{2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}
 
\end{array}</math>
 
\end{array}</math>
  
 
So, we have  
 
So, we have  
−
::<math>\int \frac{\cos(x)}{(5+\sin x)^2}~dx=-\frac{1}{5+\sin(x)}+C.</math>
+
::<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
  
−
== Exercise 3 ==
+
::<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>
  
−
Evaluate the indefinite integral &nbsp;<math style="vertical-align: -16px">\int \frac{x+5}{2x+3}~dx.</math>
+
Now, you would only need to use the Product Rule once instead of twice.
  
−
Here, the substitution is not obvious.
+
== Exercise 3 ==
  
−
Let &nbsp;<math style="vertical-align: -3px">u=2x+3.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=2~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{2}=dx.</math>
+
Calculate the derivative of &nbsp;<math style="vertical-align: -16px">h(x)=\frac{x^2\sin x+1}{x^2\cos x+3}.</math>
  
−
Now, we need a way of getting rid of &nbsp;<math style="vertical-align: -2px">x+5</math>&nbsp; in the numerator.
+
Using the Quotient Rule, we have
  
−
Solving for &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in the first equation, we get &nbsp;<math style="vertical-align: -14px">x=\frac{1}{2}u-\frac{3}{2}.</math>  
+
::<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>
  
−
Plugging these into our integral, we get
+
Now, we need to use the Product Rule. So, we have
  
 
::<math>\begin{array}{rcl}
 
::<math>\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{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{1}{2}\int \frac{\frac{1}{2}u+\frac{7}{2}}{u}~du}\\
+
& = & \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}.}
−
&&\\
 
−
& = & \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}</math>
 
\end{array}</math>
  
 
So, we get
 
So, we get
−
::<math>\int \frac{x+5}{2x+3}~dx=\frac{1}{4}(2x+3+7\ln|2x+3|)+C.</math>
+
::<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 ==
 
== Exercise 4 ==
  
−
Evaluate the indefinite integral &nbsp;<math style="vertical-align: -14px">\int \frac{x^2+4}{x+2}~dx.</math>
+
Calculate the derivative of  &nbsp;<math style="vertical-align: -14px">f(x)=\frac{e^x}{x^2\sin x}.</math>
  
−
Let &nbsp;<math style="vertical-align: -2px">u=x+2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=dx.</math>
+
First, using the Quotient Rule, we have
  
−
Now, we need a way of replacing &nbsp;<math style="vertical-align: -2px">x^2+4.</math>  
+
::<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}}\\
−
If we solve for &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in our first equation, we get &nbsp;<math style="vertical-align: -1px">x=u-2.</math>
+
&&\\
−
 
+
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
−
Now, we square both sides of this last equation to get &nbsp;<math style="vertical-align: -5px">x^2=(u-2)^2.</math>
+
\end{array}</math>
  
−
Plugging in to our integral, we get
+
Now, we need to use the Product Rule. So, we have
  
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\displaystyle{\int \frac{x^2+4}{x+2}~dx} & = & \displaystyle{\int \frac{(u-2)^2+4}{u}~du}\\
+
\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{\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.}\\
+
& = & \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>
 
\end{array}</math>
  
 
So, we have  
 
So, we have  
−
::<math>\int \frac{x^2+4}{x+2}~dx=\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.</math>
+
::<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>

Latest revision as of 22:46, 4 October 2017

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 need to know 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 \csc x.}   Recall

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 \csc x =\frac{1}{\sin x}.}

Now, 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 \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}}

Using the Product Rule and 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 \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}}

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 f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}

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.}

Notice that the function  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 the product of three functions.

We start by grouping two of the functions together. 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 g(x)=(2x\sin x)\sec x.}

Using the Product 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{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}}

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

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{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}}

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 g'(x)=2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}

But, there is another way to do this problem. 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 \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}}

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

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}.}

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 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}.}

Now, we need to use the Product Rule. 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{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}}

So, we get

Exercise 4

Calculate the derivative of  

First, using the Quotient Rule, we have

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

So, we have

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