MathAdmin: Created page with " Find the derivative of the following functions: (a) $g(\theta)=\frac{\pi^2}{(\sec\theta -\sin..."$

2017-04-10T16:28:32Z

Created page with " Find the derivative of the following functions: (a) <math style="vertical-align: -18px">g(\theta)=\frac{\pi^2}{(\sec\theta -\sin..."

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Find the derivative of the following functions:

(a)  <math style="vertical-align: -18px">g(\theta)=\frac{\pi^2}{(\sec\theta -\sin 2\theta)^2}</math>

(b)  <math style="vertical-align: -5px">y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''Chain Rule'''
|-
|        <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
|-
|'''2.''' '''Trig Derivatives'''
|-
|        <math>\frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\sec x)=\sec x \tan x</math>
|-
|'''3.''' '''Inverse Trig Derivatives
|-
|        <math>\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}</math>
|
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|       <math>g(\theta)=\pi^2(\sec\theta -\sin 2\theta)^{-2}.</math>
|-
|Now, using the Chain Rule, we have
|-
|       <math>g'(\theta)=(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}(\sec\theta -\sin 2\theta)'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Chain Rule a second time, we get
|-
|        <math>\begin{array}{rcl}
\displaystyle{g'(\theta)} & = & \displaystyle{(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}(\sec\theta -\sin 2\theta)'}\\
&&\\
& = & \displaystyle{(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}(\sec\theta\tan\theta -\cos (2\theta)(2\theta)')}\\
&&\\
& = & \displaystyle{(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}(\sec\theta\tan\theta -\cos (2\theta)(2))}\\
&&\\
& = & \displaystyle{\frac{-2\pi^2(\sec\theta\tan\theta -2\cos (2\theta))}{(\sec\theta -\sin 2\theta)^{3}}.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we have
|-
|        <math>y'=(\cos(3\pi))'+(\tan^{-1}(\sqrt{x}))'.</math>
|-
|Since  <math style="vertical-align: -5px">\cos(3\pi)</math>  is a constant,
|-
|we have
|-
|        <math>(\cos(3\pi))'=0.</math>
|-
|Therefore,
|-
|        <math>y'=(\tan^{-1}(\sqrt{x}))'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Chain Rule, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{y'} & = & \displaystyle{(\tan^{-1}(\sqrt{x}))'}\\
&&\\
& = & \displaystyle{\bigg(\frac{1}{1+(\sqrt{x})^2}\bigg)(\sqrt{x})'}\\
&&\\
& = & \displaystyle{\bigg(\frac{1}{1+x}\bigg)\frac{1}{2\sqrt{x}}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>g'(\theta)=\frac{-2\pi^2(\sec\theta\tan\theta -2\cos (2\theta))}{(\sec\theta -\sin 2\theta)^{3}}</math>
|-
|   '''(b)'''    <math>y'=\bigg(\frac{1}{1+x}\bigg)\frac{1}{2\sqrt{x}}</math>
|}
[[009A_Sample_Final_3|'''Return to Sample Exam''']]

009A Sample Final 3, Problem 2 - Revision history

MathAdmin: Created page with " Find the derivative of the following functions: (a) g(\theta)=\frac{\pi^2}{(\sec\theta -\sin..."

MathAdmin: Created page with " Find the derivative of the following functions: (a) $g(\theta)=\frac{\pi^2}{(\sec\theta -\sin..."$