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| | <span class="exam">(b) <math style="vertical-align: -5px">y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math> | | <span class="exam">(b) <math style="vertical-align: -5px">y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 3, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' '''Chain Rule'''
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| − | | <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
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| − | |- | |
| − | |'''2.''' '''Trig Derivatives'''
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| − | | <math>\frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\sec x)=\sec x \tan x</math>
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| − | |-
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| − | |'''3.''' '''Inverse Trig Derivatives
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| − | | <math>\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}</math>
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| − | |}
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| − | '''Solution:''' | + | [[009A Sample Final 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we write
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| − | | <math>g(\theta)=\pi^2(\sec\theta -\sin 2\theta)^{-2}.</math>
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| − | |-
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| − | |Now, using the Chain Rule, we have
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| − | |-
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| − | | <math>g'(\theta)=(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}(\sec\theta -\sin 2\theta)'.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, using the Chain Rule a second time, we get
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{g'(\theta)} & = & \displaystyle{(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}(\sec\theta -\sin 2\theta)'}\\
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| − | &&\\
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| − | & = & \displaystyle{(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}(\sec\theta\tan\theta -\cos (2\theta)(2\theta)')}\\
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| − | &&\\
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| − | & = & \displaystyle{(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}(\sec\theta\tan\theta -\cos (2\theta)(2))}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-2\pi^2(\sec\theta\tan\theta -2\cos (2\theta))}{(\sec\theta -\sin 2\theta)^{3}}.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we have
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| − | |-
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| − | | <math>y'=(\cos(3\pi))'+(\tan^{-1}(\sqrt{x}))'.</math>
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| − | |-
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| − | |Since <math style="vertical-align: -5px">\cos(3\pi)</math> is a constant,
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| − | |-
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| − | |we have
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| − | | <math>(\cos(3\pi))'=0.</math>
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| − | |-
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| − | |Therefore,
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| − | | <math>y'=(\tan^{-1}(\sqrt{x}))'.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, using the Chain Rule, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{y'} & = & \displaystyle{(\tan^{-1}(\sqrt{x}))'}\\
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| − | &&\\
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| − | & = & \displaystyle{\bigg(\frac{1}{1+(\sqrt{x})^2}\bigg)(\sqrt{x})'}\\
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| − | &&\\
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| − | & = & \displaystyle{\bigg(\frac{1}{1+x}\bigg)\frac{1}{2\sqrt{x}}.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math>g'(\theta)=\frac{-2\pi^2(\sec\theta\tan\theta -2\cos (2\theta))}{(\sec\theta -\sin 2\theta)^{3}}</math>
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| − | |-
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| − | | '''(b)''' <math>y'=\bigg(\frac{1}{1+x}\bigg)\frac{1}{2\sqrt{x}}</math>
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| − | |}
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| | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
