MathAdmin at 01:11, 21 May 2017

2017-05-21T01:11:49Z

MathAdmin: Created page with "Compute $\frac{dy}{dx}.$ (a) $y=\bigg(\frac{x^2+3}{x^2-1}\bigg)^3$
2017-04-10T16:21:12Z

Created page with "Compute <math>\frac{dy}{dx}.</math> (a) <math style="vertical-align: -15px">y=\bigg(\frac{x^2+3}{x^2-1}\bigg)^3</math> <sp..."

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Compute   <math>\frac{dy}{dx}.</math>

(a)  <math style="vertical-align: -15px">y=\bigg(\frac{x^2+3}{x^2-1}\bigg)^3</math>

(b)  <math style="vertical-align: -4px">y=x\cos(\sqrt{x+1})</math>

(c)  <math style="vertical-align: -5px">y=\sin^{-1} x</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''Product Rule'''
|-
|        <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
|-
|'''2.''' '''Quotient Rule'''
|-
|        <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
|-
|'''3.''' '''Chain Rule'''
|-
|        <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Chain Rule, we have
|-
|        <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>
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Quotient Rule, we have
|-
|        <math>\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}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Product Rule, we have
|-
|        <math>\frac{dy}{dx}=x(\cos(\sqrt{x+1}))'+(x)'\cos(\sqrt{x+1}).</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using the Chain Rule, we get
|-
|        <math>\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}</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -5px">f(x)=\sin(x)</math>  and  <math style="vertical-align: -5px">g(x)=\sin^{-1}x.</math>
|-
|These functions are inverses of each other since
|-
|       <math style="vertical-align: -5px">f(g(x))=x</math>  and   <math style="vertical-align: -5px">g(f(x))=x.</math>
|-
|Therefore,
|-
|        <math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{\frac{1}{f'(g(x))}}\\
&&\\
& = & \displaystyle{\frac{1}{\cos(\sin^{-1} x)}.}
\end{array}</math>
|-
|Now, let  <math style="vertical-align: -5px">y=\sin^{-1}(x).</math>  Then,  <math style="vertical-align: -5px">x=\sin(y).</math>
|-
|So,  <math style="vertical-align: -5px">\cos(\sin^{-1} x)=\cos(y).</math>
|-
|Therefore,
|-
|       <math>g'(x)=\frac{1}{\cos(y)}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, since
|-
|       <math>\cos^2 y+\sin^2 y =1,</math>
|-
|we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\cos(y)} & = & \displaystyle{\sqrt{1-\sin^2 y}}\\
&&\\
& = & \displaystyle{\sqrt{1-x^2}.}
\end{array}</math>
|-
|Hence,
|-
|       <math>g'(x)=\frac{1}{\sqrt{1-x^2}}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\frac{3(x^2+3)^2(-8x)}{(x^2-1)^4}</math>
|-
|   '''(b)'''    <math>\frac{-x\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\cos(\sqrt{x+1})</math>
|-
|   '''(c)'''    <math>\frac{1}{\sqrt{1-x^2}}</math>
|}
[[009A_Sample_Final_2|'''Return to Sample Exam''']]

@@ Line 88: / Line 88: @@
 !Step 1: &nbsp;
 |-
-|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>
+|Let &nbsp;<math style="vertical-align: -5px">y=\sin^{-1}(x).</math>&nbsp; Then,
 |-
-|These functions are inverses of each other since
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sin(y)=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>
+|for &nbsp;<math>y</math>&nbsp; in the interval &nbsp;<math>\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].</math>
 |-
-|Therefore,
+|Using implicit differentiation, we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos(y) \frac{dy}{dx}=1.</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>
+|Solving for &nbsp;<math style="vertical-align: -15px">\frac{dy}{dx},</math>&nbsp; we get
 |-
-|So, &nbsp;<math style="vertical-align: -5px">\cos(\sin^{-1} x)=\cos(y).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{1}{\cos(y)}.</math>
 |}
@@ Line 114: / Line 106: @@
 !Step 2: &nbsp;
 |-
-|Now, since
+|Now, since &nbsp;<math>\sin(y)=x,</math>&nbsp; we have the following diagram.
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos^2 y+\sin^2 y =1,</math>
+|(Insert diagram)
 |-
-|we have
+|Therefore,
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\cos(y)} & = & \displaystyle{\sqrt{1-\sin^2 y}}\\
+\displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{1}{\cos(y)}}\\
 &&\\
-& = & \displaystyle{\sqrt{1-x^2}.}
+& = & \displaystyle{\frac{1}{\sqrt{1-x^2}}.}
 \end{array}</math>
 |}

009A Sample Final 2, Problem 3 - Revision history

MathAdmin at 01:11, 21 May 2017