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| − | <span class="exam">Find the derivatives of the following functions. Do not simplify. | + | <span class="exam"> Let <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math> |
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| − | <span class="exam">(a) <math style="vertical-align: -5px">f(x)=\sqrt{x}(x^2+2)</math> | + | <span class="exam">(a) Use the definition of the derivative to compute <math>\frac{dy}{dx}</math> for <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math> |
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| − | <span class="exam">(b) <math style="vertical-align: -17px">g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math style="vertical-align: 0px">x>0</math> | + | <span class="exam">(b) Find the equation of the tangent line to <math style="vertical-align: -5px">y=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1).</math> |
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| − | <span class="exam">(c) <math style="vertical-align: -20px">h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math>
| + | (insert picture of handwritten solution) |
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| | + | [[009A Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |'''1.''' '''Product Rule'''
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| − | |-
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| − | | <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
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| − | |-
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| − | |'''2.''' '''Quotient Rule'''
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| − | |-
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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.''' '''Chain Rule'''
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| − | |-
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| − | | <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
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| − | |}
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| − |
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| − |
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| − | '''Solution:'''
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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 Product Rule, we have
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| − | |-
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| − | | <math>f'(x)=(\sqrt{x})'(x^2+2)+\sqrt{x}(x^2+2)'.</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 have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{f'(x)} & = & \displaystyle{(\sqrt{x})'(x^2+2)+\sqrt{x}(x^2+2)'}\\
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| − | &&\\
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| − | & = & \displaystyle{\bigg(\frac{1}{2}x^{-\frac{1}{2}}\bigg)(x^2+2)+\sqrt{x}(2x).}
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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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| − | |Using the Quotient Rule, we have
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| − | |-
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| − | | <math>g'(x)=\frac{(x^{\frac{3}{2}}+2)(x+3)'-(x+3)(x^{\frac{3}{2}}+2)'}{(x^{\frac{3}{2}}+2)^2}.</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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| − | |Now, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{g'(x)} & = & \displaystyle{\frac{(x^{\frac{3}{2}}+2)(x+3)'-(x+3)(x^{\frac{3}{2}}+2)'}{(x^{\frac{3}{2}}+2)^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(c)'''
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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 Quotient Rule, we have
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| − | |-
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| − | | <math>h'(x)=\frac{\sqrt{x^2+1}(e^{-5x^3})'-e^{-5x^3}(\sqrt{x^2+1})'}{(\sqrt{x^2+1})^2}.</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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| − | |Now, using the Chain Rule, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{h'(x)} & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})'-e^{-5x^3}(\sqrt{x^2+1})'}{(\sqrt{x^2+1})^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})(-5x^3)'-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(x^2+1)'}{(\sqrt{x^2+1})^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})(-15x^2)-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(2x)}{(\sqrt{x^2+1})^2}.}
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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>\bigg(\frac{1}{2}x^{-\frac{1}{2}}\bigg)(x^2+2)+\sqrt{x}(2x)</math>
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| − | |-
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| − | | '''(b)''' <math>\frac{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}</math>
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| − | |-
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| − | | '''(c)''' <math>\frac{\sqrt{x^2+1}(e^{-5x^3})(-15x^2)-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(2x)}{(\sqrt{x^2+1})^2}</math>
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| − | |}
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| | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
Let 
(a) Use the definition of the derivative to compute 
for
(b) Find the equation of the tangent line to 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=\sqrt{3x-5}}
at 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 (2,1).}
(insert picture of handwritten solution)
Detailed Solution for this Problem
Return to Sample Exam