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| | <span class="exam">(b) <math style="vertical-align: -3px">g(x)=2\sin (4x)+4\tan (\sqrt{1+x^3})</math> | | <span class="exam">(b) <math style="vertical-align: -3px">g(x)=2\sin (4x)+4\tan (\sqrt{1+x^3})</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 1, Problem 3 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.''' '''Quotient Rule'''
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| − | | <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
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| − | |-
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| − | |'''3.''' '''Trig Derivatives'''
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| − | | <math>\frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\tan x)=\sec^2 x</math>
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| − | |}
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| − | '''Solution:''' | + | [[009A Sample Final 1, Problem 3 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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| − | |Using the Chain Rule, we have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{\bigg(\frac{x^2-1}{x^2+1}\bigg)}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).}\\
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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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| − | !Step 2:
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| − | |-
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| − | |Now, we need to calculate <math>\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).</math>
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| − | |-
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| − | |To do this, we use the Quotient Rule. So, we have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{4x}{(x^2+1)^2}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4x}{(x^2-1)(x^2+1)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4x}{x^4-1}.}\\
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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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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| − | |We need to use the Chain Rule. We have
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| − | |-
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| − | <math>g'(x)\,=\,8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg).</math>
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| − | |}
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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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| − | |We need to calculate <math>\frac{d}{dx}\sqrt{1+x^3}.</math>
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| − | |-
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| − | |We use the Chain Rule again to get
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{g'(x)} & = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\frac{1}{2}(1+x^3)^{-\frac{1}{2}}3x^2}\\
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| − | &&\\
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| − | & = & \displaystyle{8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}.}\\
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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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| − | | '''(a)''' <math style="vertical-align: -14px">f'(x)=\frac{4x}{x^4-1}</math>
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| − | |-
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| − | | '''(b)''' <math style="vertical-align: -18px">g'(x)=8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}</math>
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| − | |}
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| | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
