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

Revision as of 18:36, 7 March 2017

Compute ${\frac {dy}{dx}}.$

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

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

(c) $y=\sin ^{-1}x$

Foundations:
1. Product Rule
${\frac {d}{dx}}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)$
2. Quotient Rule
${\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}$
3. Chain Rule
${\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$

Solution:

(a)

Step 1:
Using the Chain Rule, we have
${\frac {dy}{dx}}=3{\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}^{2}{\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}'.$

Step 2:

Now, using the Quotient Rule, we have

{\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:
Using the Product Rule, we have
${\frac {dy}{dx}}=x(\cos({\sqrt {x+1}}))'+(x)'\cos({\sqrt {x+1}}).$

Step 2:

Now, using the Chain Rule, we get

{\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 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(x)} and 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)=\sin^{-1}x.}
These functions are inverses of each other since
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(g(x))=x} and 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(f(x))=x.}
Therefore,
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}{f'(g(x))}}\\ &&\\ & = & \displaystyle{\frac{1}{\cos(\sin^{-1} x)}.} \end{array}}
Now, let 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).} Then, 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=\sin(y).}
So, 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 \cos(\sin^{-1} x)=\cos(y).}
Therefore,
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}{\cos(y)}.}

Step 2:
Now, since
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 \cos^2 y+\sin^2 y =1,}
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{\cos(y)} & = & \displaystyle{\sqrt{1-\sin^2 y}}\\ &&\\ & = & \displaystyle{\sqrt{1-x^2}.} \end{array}}
Hence,
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}{\sqrt{1-x^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) 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\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\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 \frac{1}{\sqrt{1-x^2}}}

Return to Sample Exam

@@ Line 88: / Line 88: @@
 !Step 1: &nbsp;
 |-
-|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
 |-
-|&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,
@@ Line 102: / Line 102: @@
 \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,
@@ Line 117: / Line 117: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos^2 y+\sin^2 y =1,</math>
+|-
+|we have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
@@ Line 137: / Line 139: @@
 |&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>''']]

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

Revision as of 18:36, 7 March 2017

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