Difference between revisions of "009A Sample Midterm 3, Problem 6"

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!Step 1:    
 
!Step 1:    
 
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|First, using the Chain Rule, we have
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|&nbsp; &nbsp; &nbsp; &nbsp; <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>
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Now, using the Quotient Rule, we have
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\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)'}\\
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&&\\
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& = & \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)}\\
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&&\\
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& = & \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).}
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\end{array}</math>
 
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|'''(a)'''  
 
|'''(a)'''  
 
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|'''(b)'''
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|&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>
 
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|&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>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|'''<u>Return to Sample Exam</u>''']]

Revision as of 16:04, 17 February 2017

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

a)Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\sin {\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}}
b)Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)={\sqrt {\frac {x^{2}+2}{x^{2}+4}}}}
c)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 h(x)=(x+\cos^2x)^8}


Foundations:  
1. Chain Rule
2. Quotient Rule


Solution:

(a)

Step 1:  
Step 2:  

(b)

Step 1:  
First, using the Chain Rule, we have
        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 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)'.}
Step 2:  
Now, using the Quotient Rule, we have

        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 \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}}

(c)

Step 1:  
First, using the Chain Rule, we have
        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 h'(x)=8(x+\cos^2(x))^7(x+\cos^2(x))'.}
Step 2:  
Now, using the Chain Rule again we get

        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 \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}}


Final Answer:  
(a)
    (b)     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{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)}
    (c)     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 8(x+\cos^2(x))^7(1-2\cos(x)\sin(x))}

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