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

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
|Let &nbsp;<math>f(x)=\sin(x)</math>&nbsp; and &nbsp;<math>g(x)=\sin^{-1}x.</math>
+
|Let &nbsp;<math style="vertical-align: -5px">f(x)=\sin(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g(x)=\sin^{-1}x.</math>
 
|-
 
|-
 
|These functions are inverses of each other since
 
|These functions are inverses of each other since
 
|-
 
|-
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(g(x))=x</math>&nbsp; and &nbsp; <math>g(f(x))=x.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(g(x))=x</math>&nbsp; and &nbsp; <math style="vertical-align: -5px">g(f(x))=x.</math>
 
|-
 
|-
 
|Therefore,
 
|Therefore,
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\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
|Now, let &nbsp;<math>y=\sin^{-1}(x).</math>&nbsp; Then, &nbsp;<math>x=\sin(y).</math>
+
|Now, let &nbsp;<math style="vertical-align: -5px">y=\sin^{-1}(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">x=\sin(y).</math>
 
|-
 
|-
|So, &nbsp;<math>\cos(\sin^{-1} x)=\cos(y).</math>
+
|So, &nbsp;<math style="vertical-align: -5px">\cos(\sin^{-1} x)=\cos(y).</math>
 
|-
 
|-
 
|Therefore,  
 
|Therefore,  
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|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos^2 y+\sin^2 y =1,</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos^2 y+\sin^2 y =1,</math>
 +
|-
 +
|we have
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{-x\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\cos(\sqrt{x+1})</math>
 
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{-x\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\cos(\sqrt{x+1})</math>
 
|-
 
|-
|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>g'(x)=\frac{1}{\sqrt{1-x^2}}</math>
+
|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>\frac{1}{\sqrt{1-x^2}}</math>
 
|}
 
|}
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 18:36, 7 March 2017

Compute  

(a)  

(b)  

(c)  

Foundations:  
1. Product Rule
       
2. Quotient Rule
       
3. Chain Rule
       


Solution:

(a)

Step 1:  
Using the Chain Rule, we have
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
       

(b)

Step 1:  
Using the Product Rule, we have
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {dy}{dx}}=x(\cos({\sqrt {x+1}}))'+(x)'\cos({\sqrt {x+1}}).}
Step 2:  
Now, using the Chain Rule, we get
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\frac {dy}{dx}}&=&\displaystyle {x(\cos({\sqrt {x+1}}))'+(x)'\cos({\sqrt {x+1}})}\\&&\\&=&\displaystyle {x(-\sin({\sqrt {x+1}}))({\sqrt {x+1}})'+(1)\cos({\sqrt {x+1}})}\\&&\\&=&\displaystyle {-x\sin({\sqrt {x+1}}){\frac {1}{2{\sqrt {x+1}}}}(x+1)'+\cos({\sqrt {x+1}})}\\&&\\&=&\displaystyle {{\frac {-x\sin({\sqrt {x+1}})}{2{\sqrt {x+1}}}}+\cos({\sqrt {x+1}}).}\end{array}}}

(c)

Step 1:  
Let    and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=\sin ^{-1}x.}
These functions are inverses of each other since
         and  
Therefore,
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\frac {1}{f'(g(x))}}\\&&\\&=&\displaystyle {{\frac {1}{\cos(\sin ^{-1}x)}}.}\end{array}}}
Now, let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=\sin ^{-1}(x).}   Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=\sin(y).}
So,  
Therefore,
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g'(x)={\frac {1}{\cos(y)}}.}
Step 2:  
Now, since
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \cos ^{2}y+\sin ^{2}y=1,}
we have
       
Hence,
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g'(x)={\frac {1}{\sqrt {1-x^{2}}}}.}


Final Answer:  
   (a)    Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {3(x^{2}+3)^{2}(-8x)}{(x^{2}-1)^{4}}}}
   (b)   
   (c)    Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{\sqrt {1-x^{2}}}}}

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