View source for 007A Sample Final 1, Problem 5 Detailed Solution

<span class="exam"> If &nbsp;<math style="vertical-align: -5px">y=\cos^{-1} (2x)</math> compute &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; and find the equation for the tangent line at &nbsp;<math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}.</math>

<span class="exam">You may leave your answers in point-slope form.
<hr>
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!Background Information: &nbsp; 
|-
|'''1.''' '''Chain Rule'''
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
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|'''2.''' Recall
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{d}{dx}(\cos^{-1}(x))=\frac{-1}{\sqrt{1-x^2}}</math>
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|'''3.''' The equation of the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; is
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">m=f'(a).</math>
|}


'''Solution:'''

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!Step 1: &nbsp; 
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|First, we compute &nbsp;<math style="vertical-align: -13px">\frac{dy}{dx}.</math>
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|Using the Chain Rule, we get
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|
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
\displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{-1}{\sqrt{1-(2x)^2}}(2x)'}\\
&&\\
& = & \displaystyle{\frac{-2}{\sqrt{1-4x^2}}.}
\end{array}</math>
|}

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!Step 2: &nbsp;
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|To find the equation of the tangent line, we first find the slope of the line. 
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|Using &nbsp;<math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}</math>&nbsp; in the formula for &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; from Step 1, we get
|-
|
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
\displaystyle{m} & = & \displaystyle{\frac{-2}{\sqrt{1-4(\frac{\sqrt{3}}{4})^2}}}\\
&&\\
& = & \displaystyle{\frac{-2}{\sqrt{\frac{1}{4}}}}\\
&&\\
& = & \displaystyle{-4.}
\end{array}</math>
|}

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!Step 3: &nbsp;
|-
|To get a point on the line, we plug in &nbsp;<math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}</math>&nbsp; into the equation given. 
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|So, we have
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|
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
\displaystyle{y_0} & = & \displaystyle{\cos^{-1}\bigg(2\frac{\sqrt{3}}{4}\bigg)}\\
&&\\
& = & \displaystyle{\cos^{-1}\bigg(\frac{\sqrt{3}}{2}\bigg)}\\
&&\\
& = & \displaystyle{\frac{\pi}{6}.}
\end{array}</math>
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|Thus, the equation of the tangent line is &nbsp; <math style="vertical-align: -14px">y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}.</math>
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!Final Answer: &nbsp; 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{-2}{\sqrt{1-4x^2}}</math>
|-
|
&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}</math>
|}
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