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| | <span class="exam"> Find the derivatives of the following functions. Do not simplify. | | <span class="exam"> Find the derivatives of the following functions. Do not simplify. |
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| − | <span class="exam">(a) <math style="vertical-align: -16px">f(x)=\frac{(3x-5)(-x^{-2}+4x)}{x^{\frac{4}{5}}}</math> | + | <span class="exam">(a) <math style="vertical-align: -16px">f(x)=\sin\bigg(\frac{x^{-3}}{e^{-x}}\bigg)</math> |
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| − | <span class="exam">(b) <math>g(x)=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{\pi}</math> for <math style="vertical-align: 0px">x>0.</math> | + | <span class="exam">(b) <math style="vertical-align: -18px">g(x)=\sqrt{\frac{x^2+2}{x^2+4}}</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <span class="exam">(c) <math style="vertical-align: -6px">h(x)=(x+\cos^2x)^8</math> |
| − | !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.''' '''Power Rule'''
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| − | |-
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| − | | <math>\frac{d}{dx}(x^n)=nx^{n-1}</math>
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| − | |}
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| | + | <hr> |
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| − | '''Solution:''' | + | [[009A Sample Midterm 3, Problem 5 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']] |
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| − | '''(a)'''
| + | [[File:9ASM3P5.jpg|600px|thumb|center]] |
| − | {| 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 | |
| − | |- | |
| − | | <math>f'(x)=\frac{x^{\frac{4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac{4}{5}})'}{(x^{\frac{4}{5}})^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 use the Product Rule to get
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{f'(x)} & = & \displaystyle{\frac{x^{\frac{4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac{4}{5}})'}{(x^{\frac{4}{5}})^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(-x^{-2}+4x)'+(3x-5)'(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{-\frac{1}{5}})}{(x^{\frac{4}{5}})^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{-\frac{1}{5}})}{(x^{\frac{4}{5}})^2}.}
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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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| − | |First, we have
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| − | |-
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| − | | <math>g'(x)=(\sqrt{x})'+\bigg(\frac{1}{\sqrt{x}}\bigg)'+(\sqrt{\pi})'.</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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| − | |Since <math style="vertical-align: 0px">\pi</math> is a constant, <math style="vertical-align: -3px">\sqrt{\pi}</math> is also a constant.
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| − | |-
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| − | |Hence,
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| − | |-
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| − | | <math>(\sqrt{\pi})'=0.</math>
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| − | |-
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| − | |Therefore, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{g'(x)} & = & \displaystyle{(\sqrt{x})'+\bigg(\frac{1}{\sqrt{x}}\bigg)'+(\sqrt{\pi})'}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}+0}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{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>f'(x)=\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{-\frac{1}{5}})}{(x^{\frac{4}{5}})^2}</math>
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| − | |-
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| − | | '''(b)''' <math>g'(x)=\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}</math>
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| − | |}
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| | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
