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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> |
| | + | <hr> |
| | + | [[009A Sample Midterm 1, Problem 4 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']] |
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| − | <span class="exam">(c) <math style="vertical-align: -20px">h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math>
| + | [[File:9ASM1P4.jpg|600px|thumb|center]] |
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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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| − | '''Solution:'''
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| − |
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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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| − | {| 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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| − |
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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>f'(x)=\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>g'(x)=\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>h'(x)=\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 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}}
for 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}.}
(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).}
Detailed Solution with Background Information
Return to Sample Exam
