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

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(Created page with "<span class="exam"> Find the derivatives of the following functions. Do not simplify. <span class="exam">(a)  <math style="vertical-align: -16px">f(x)=\sin\bigg(\frac{x^...")
 
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&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>  
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|&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>  
 
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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; '''(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>
 
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|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>8(x+\cos^2(x))^7(1-2\cos(x)\sin(x))</math>
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|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>h'(x)=8(x+\cos^2(x))^7(1-2\cos(x)\sin(x))</math>
 
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[[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 18:49, 13 April 2017

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

(a)  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 f(x)=\sin\bigg(\frac{x^{-3}}{e^{-x}}\bigg)}

(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 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
        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{d}{dx}(f(g(x)))=f'(g(x))g'(x)}
2. Quotient Rule
        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{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}}


Solution:

(a)

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 f'(x)=\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{x^{-3}}{e^{-x}}\bigg)'.}
Step 2:  
Now, using the Quotient Rule and 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 \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}}

(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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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)     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 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)}
    (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 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)}
    (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)=8(x+\cos^2(x))^7(1-2\cos(x)\sin(x))}

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