Difference between revisions of "009A Sample Midterm 2, Problem 4"

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!Step 1:    
 
!Step 1:    
 
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|Using the Quotient Rule, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>g'(x)=\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^2}.</math>
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Now, we have
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{g'(x)} & = & \displaystyle{\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^2}}\\
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&&\\
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& = & \displaystyle{\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}.}
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\end{array}</math>
 
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|&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>x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1)</math>  
 
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|'''(b)'''  
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}</math>
 
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[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 12:24, 17 February 2017

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

a)
b) where 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 x>0}


Foundations:  
1. Product Rule
2. Quotient Rule


Solution:

(a)

Step 1:  
Using the Product 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)=x^3(x^{\frac{4}{3}}-1)'+(x^3)'(x^{\frac{4}{3}}-1).}
Step 2:  
Now, 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{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}}

(b)

Step 1:  
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 g'(x)=\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^2}.}
Step 2:  
Now, 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+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}}


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 x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1)}
    (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+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}}

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