Difference between revisions of "009A Sample Final 2, Problem 3"

Revision as of 09:10, 1 December 2017

Compute 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{dy}{dx}.}

(a) 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 y=\bigg(\frac{x^2+3}{x^2-1}\bigg)^3}

(b) 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 y=x\cos(\sqrt{x+1})}

(c) 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 y=\sin^{-1} x}

Solution

Detailed Solution

Return to Sample Exam

@@ Line 7: / Line 7: @@
 <span class="exam">(c) &nbsp;<math style="vertical-align: -5px">y=\sin^{-1} x</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Midterm 2, Problem 3 Solution|'''<u>Solution</u>''']]
-|-
-|'''1.''' '''Product Rule'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
-|-
-|'''2.''' '''Quotient Rule'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
-|-
-|'''3.''' '''Chain Rule'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
-|}
-'''Solution:'''
+[[009A Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
-!Step 1: &nbsp;
-|-
-|Using the Chain Rule, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{x^2+3}{x^2-1}\bigg)'.</math>
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, using the Quotient Rule, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\frac{dy}{dx}} & = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{x^2+3}{x^2-1}\bigg)'}\\
-&&\\
-& = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{(x^2-1)(x^2+3)'-(x^2+3)(x^2-1)'}{(x^2-1)^2}\bigg)}\\
-&&\\
-& = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{(x^2-1)(2x)-(x^2+3)(2x)}{(x^2-1)^2}\bigg)}\\
-&&\\
-& = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{2x^3-2x-2x^3-6x}{(x^2-1)^2}\bigg)}\\
-&&\\
-& = & \displaystyle{\frac{3(x^2+3)^2(-8x)}{(x^2-1)^4}.}
-\end{array}</math>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Using the Product Rule, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=x(\cos(\sqrt{x+1}))'+(x)'\cos(\sqrt{x+1}).</math>
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, using the Chain Rule, we get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\frac{dy}{dx}} & = & \displaystyle{x(\cos(\sqrt{x+1}))'+(x)'\cos(\sqrt{x+1})}\\
-&&\\
-& = & \displaystyle{x(-\sin(\sqrt{x+1}))(\sqrt{x+1})'+(1)\cos(\sqrt{x+1})}\\
-&&\\
-& = & \displaystyle{-x\sin(\sqrt{x+1})\frac{1}{2\sqrt{x+1}}(x+1)'+\cos(\sqrt{x+1})}\\
-&&\\
-& = & \displaystyle{\frac{-x\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\cos(\sqrt{x+1}).}
-\end{array}</math>
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Let &nbsp;<math style="vertical-align: -5px">y=\sin^{-1}(x).</math>&nbsp; Then,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sin(y)=x</math>
-|-
-|for &nbsp;<math>y</math>&nbsp; in the interval &nbsp;<math>\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].</math>
-|-
-|Using implicit differentiation, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos(y) \frac{dy}{dx}=1.</math>
-|-
-|Solving for &nbsp;<math style="vertical-align: -15px">\frac{dy}{dx},</math>&nbsp; we get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{1}{\cos(y)}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, since &nbsp;<math>\sin(y)=x,</math>&nbsp; we have the following diagram.
-|-
-|(Insert diagram)
-|-
-|Therefore,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\cos(y)=\sqrt{1-x^2}.</math>
-|-
-|Hence,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{1}{\cos(y)}}\\
-&&\\
-& = & \displaystyle{\frac{1}{\sqrt{1-x^2}}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{3(x^2+3)^2(-8x)}{(x^2-1)^4}</math>
-|-
-|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{-x\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\cos(\sqrt{x+1})</math>
-|-
-|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>\frac{1}{\sqrt{1-x^2}}</math>
-|}
-[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Final 2, Problem 3"

Revision as of 09:10, 1 December 2017

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