MathAdmin at 02:49, 14 April 2017

2017-04-14T02:49:57Z

MathAdmin: Created page with " Find the derivatives of the following functions. Do not simplify. (a) $f(x)=\sin\bigg(\frac{x^..."$

2017-04-09T18:57:33Z

Created page with " Find the derivatives of the following functions. Do not simplify. (a) <math style="vertical-align: -16px">f(x)=\sin\bigg(\frac{x^..."

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Find the derivatives of the following functions. Do not simplify.

(a)  <math style="vertical-align: -16px">f(x)=\sin\bigg(\frac{x^{-3}}{e^{-x}}\bigg)</math>

(b)  <math style="vertical-align: -18px">g(x)=\sqrt{\frac{x^2+2}{x^2+4}}</math>

(c)  <math style="vertical-align: -6px">h(x)=(x+\cos^2x)^8</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''Chain Rule'''
|-
|        <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
|-
|'''2.''' '''Quotient Rule'''
|-
|        <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, using the Chain Rule, we have
|-
|        <math>f'(x)=\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{x^{-3}}{e^{-x}}\bigg)'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Quotient Rule and Chain Rule, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{x^{-3}}{e^{-x}}\bigg)'}\\
&&\\
& = & \displaystyle{\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(x^{-3})'-x^{-3}(e^{-x})'}{(e^{-x})^2}\bigg)}\\
&&\\
& = & \displaystyle{\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-x)'}{(e^{-x})^2}\bigg)}\\
&&\\
& = & \displaystyle{\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-1)}{(e^{-x})^2}\bigg).}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, using the Chain Rule, we have
|-
|        <math>g'(x)=\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{x^2+2}{x^2+4}\bigg)'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Quotient Rule, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{x^2+2}{x^2+4}\bigg)'}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(x^2+2)'-(x^2+2)(x^2+4)'}{(x^2+4)^2}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(2x)-(x^2+2)(2x)}{(x^2+4)^2}\bigg).}
\end{array}</math>
|-
|
|-
|
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, using the Chain Rule, we have
|-
|        <math>h'(x)=8(x+\cos^2(x))^7(x+\cos^2(x))'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Chain Rule again we get
|-
|
        <math>\begin{array}{rcl}
\displaystyle{h'(x)} & = & \displaystyle{8(x+\cos^2(x))^7(x+\cos^2(x))'}\\
&&\\
& = & \displaystyle{8(x+\cos^2(x))^7(1+2\cos(x)(\cos(x))')}\\
&&\\
& = & \displaystyle{8(x+\cos^2(x))^7(1-2\cos(x)\sin(x)).}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-1)}{(e^{-x})^2}\bigg)</math>
|-
|    '''(b)'''     <math>\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(2x)-(x^2+2)(2x)}{(x^2+4)^2}\bigg)</math>
|-
|    '''(c)'''     <math>8(x+\cos^2(x))^7(1-2\cos(x)\sin(x))</math>
|}
[[009A_Sample_Midterm_3|'''Return to Sample Exam''']]

@@ Line 104: / Line 104: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-1)}{(e^{-x})^2}\bigg)</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>f'(x)=\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-1)}{(e^{-x})^2}\bigg)</math>
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(2x)-(x^2+2)(2x)}{(x^2+4)^2}\bigg)</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>g'(x)=\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(2x)-(x^2+2)(2x)}{(x^2+4)^2}\bigg)</math>
 |-
-|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>8(x+\cos^2(x))^7(1-2\cos(x)\sin(x))</math>
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>h'(x)=8(x+\cos^2(x))^7(1-2\cos(x)\sin(x))</math>
 |}
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

009A Sample Midterm 3, Problem 6 - Revision history

MathAdmin at 02:49, 14 April 2017

MathAdmin: Created page with " Find the derivatives of the following functions. Do not simplify. (a) f(x)=\sin\bigg(\frac{x^..."

MathAdmin: Created page with " Find the derivatives of the following functions. Do not simplify. (a) $f(x)=\sin\bigg(\frac{x^..."$