Difference between revisions of "009A Sample Midterm 3, Problem 5"

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!Step 2:  
 
!Step 2:  
 
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|-
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|Now, we use the Product Rule to get
 
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|-
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{f'(x)} & = & \displaystyle{\frac{x^{\frac{4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac{4}{5}})'}{(x^{\frac{4}{5}})^2}}\\
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&&\\
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& = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(-x^{-2}+4x)'+(3x-5)'(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}}\\
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&&\\
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& = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}.}
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\end{array}</math>
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)'''  
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}</math>
 
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|'''(b)'''  
 
|'''(b)'''  
 
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[[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 15:37, 17 February 2017

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

a)
b) for 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. Quotient Rule
2. Product Rule
3. Power Rule


Solution:

(a)

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 f'(x)=\frac{x^{\frac{4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac{4}{5}})'}{(x^{\frac{4}{5}})^2}.}
Step 2:  
Now, we use the Product Rule to 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{f'(x)} & = & \displaystyle{\frac{x^{\frac{4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac{4}{5}})'}{(x^{\frac{4}{5}})^2}}\\ &&\\ & = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(-x^{-2}+4x)'+(3x-5)'(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}}\\ &&\\ & = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}.} \end{array}}

(b)

Step 1:  
Step 2:  


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 \frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}}
(b)

Return to Sample Exam