009A Sample Midterm 2, Problem 4 - Revision history

MathAdmin at 19:08, 11 November 2017

2017-11-11T19:08:29Z

MathAdmin at 02:50, 14 April 2017

2017-04-14T02:50:56Z

MathAdmin: Created page with "Find the derivatives of the following functions. Do not simplify. (a) $f(x)=x^3(x^{\frac{4}{3}}..."$

2017-04-09T18:52:42Z

Created page with "Find the derivatives of the following functions. Do not simplify. (a) <math style="vertical-align: -5px">f(x)=x^3(x^{\frac{4}{3}}..."

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

(a)   <math style="vertical-align: -5px">f(x)=x^3(x^{\frac{4}{3}}-1)</math>

(b)   <math style="vertical-align: -14px">g(x)=\frac{x^3+x^{-3}}{1+6x}</math>  where  <math style="vertical-align: 0px">x>0</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''Product Rule'''
|-
|        <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(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>
|-
|'''3.''' '''Power Rule'''
|-
|        <math>\frac{d}{dx}(x^n)=nx^{n-1}</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Product Rule, we have
|-
|        <math>f'(x)=x^3(x^{\frac{4}{3}}-1)'+(x^3)'(x^{\frac{4}{3}}-1).</math>
|}

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

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Quotient Rule, we have
|-
|        <math>g'(x)=\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^2}}\\
&&\\
& = & \displaystyle{\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1)</math>
|-
|    '''(b)'''     <math>\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}</math>
|}
[[009A_Sample_Midterm_2|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Find the derivatives of the following functions. Do not simplify.
+<span class="exam"> To determine drug dosages, doctors estimate a person's body surface area (BSA) (in meters squared) using the formula:
-<span class="exam">(a) &nbsp; <math style="vertical-align: -5px">f(x)=x^3(x^{\frac{4}{3}}-1)</math>
+::<span class="exam"><math>\text{BSA}=\frac{\sqrt{hm}}{60}</math>
-<span class="exam">(b) &nbsp; <math style="vertical-align: -14px">g(x)=\frac{x^3+x^{-3}}{1+6x}</math>&nbsp; where &nbsp;<math style="vertical-align: 0px">x>0</math>
+<span class="exam">where &nbsp;<math style="vertical-align: 0px">h</math>&nbsp; is the height in centimeters and &nbsp;<math style="vertical-align: 0px">m</math>&nbsp; is the mass in kilograms. Calculate the rate of change of BSA with respect to height for a person of a constant mass of &nbsp;<math style="vertical-align: 0px">m=85.</math>&nbsp; What is the rate at &nbsp;<math style="vertical-align: -1px">h=170</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">h=190?</math>&nbsp; Express your results in the correct units. Does the BSA increase more rapidly with respect to height at lower or higher heights?
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 72: / Line 72: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1)</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>f'(x)=x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1)</math>
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>g'(x)=\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}</math>
 |}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

009A Sample Midterm 2, Problem 4 - Revision history

MathAdmin at 19:08, 11 November 2017

MathAdmin at 02:50, 14 April 2017

MathAdmin: Created page with "Find the derivatives of the following functions. Do not simplify. (a) f(x)=x^3(x^{\frac{4}{3}}..."

MathAdmin: Created page with "Find the derivatives of the following functions. Do not simplify. (a) $f(x)=x^3(x^{\frac{4}{3}}..."$