Difference between revisions of "009A Sample Final 2, Problem 3"

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!Step 1:    
 
!Step 1:    
 
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|Using the Chain Rule, we have
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{x^2+3}{x^2-1}\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{\frac{dy}{dx}} & = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{x^2+3}{x^2-1}\bigg)'}\\
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&&\\
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& = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{(x^2-1)(x^2+3)'-(x^2+3)(x^2-1)'}{(x^2-1)^2}\bigg)}\\
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&&\\
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& = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{(x^2-1)(2x)-(x^2+3)(2x)}{(x^2-1)^2}\bigg)}\\
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&&\\
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& = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{2x^3-2x-2x^3-6x}{(x^2-1)^2}\bigg)}\\
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&&\\
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& = & \displaystyle{\frac{3(x^2+3)^2(-8x)}{(x^2-1)^4}.}
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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{3(x^2+3)^2(-8x)}{(x^2-1)^4}</math>
 
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|'''(b)'''
 
|'''(b)'''

Revision as of 17:16, 7 March 2017

Compute   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {dy}{dx}}.}

(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 y=\bigg(\frac{x^2+3}{x^2-1}\bigg)^3}

(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 y=x\cos(\sqrt{x+1})}

(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 y=\sin^{-1} x}

Foundations:  
1. Product 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(x)g(x))=f(x)g'(x)+f'(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}}
3. 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)}


Solution:

(a)

Step 1:  
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 \frac{dy}{dx}=3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{x^2+3}{x^2-1}\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{\frac{dy}{dx}} & = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{x^2+3}{x^2-1}\bigg)'}\\ &&\\ & = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{(x^2-1)(x^2+3)'-(x^2+3)(x^2-1)'}{(x^2-1)^2}\bigg)}\\ &&\\ & = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{(x^2-1)(2x)-(x^2+3)(2x)}{(x^2-1)^2}\bigg)}\\ &&\\ & = & \displaystyle{3\bigg(\frac{x^2+3}{x^2-1}\bigg)^2\bigg(\frac{2x^3-2x-2x^3-6x}{(x^2-1)^2}\bigg)}\\ &&\\ & = & \displaystyle{\frac{3(x^2+3)^2(-8x)}{(x^2-1)^4}.} \end{array}}

(b)

Step 1:  
Step 2:  

(c)

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{3(x^2+3)^2(-8x)}{(x^2-1)^4}}
(b)
(c)

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