009A Sample Final 1, Problem 3 - Revision history

MathAdmin at 16:08, 10 April 2017

2017-04-10T16:08:52Z

MathAdmin at 22:13, 18 April 2016

2016-04-18T22:13:49Z

MathAdmin at 19:03, 18 April 2016

2016-04-18T19:03:36Z

MathAdmin: Created page with "Find the derivatives of the following functions. a) $f(x)=\ln \bigg(\frac{x^2-1}{x^2+1}\bigg)$ ..."

2016-03-05T07:07:10Z

Created page with "Find the derivatives of the following functions. a) <math style="vertical-align: -16px">f(x)=\ln \bigg(\frac{x^2-1}{x^2+1}\bigg)</math>..."

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Find the derivatives of the following functions.

a) <math style="vertical-align: -16px">f(x)=\ln \bigg(\frac{x^2-1}{x^2+1}\bigg)</math>

b) <math style="vertical-align: -5px">g(x)=2\sin (4x)+4\tan (\sqrt{1+x^3})</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|For functions <math style="vertical-align: -5px">f(x)</math>  and <math style="vertical-align: -5px">g(x)</math>, recall
|-
| 
|-
|'''Chain Rule:'''  <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
|-
| 
|-
|'''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>
|-
| 
|-
|'''Trig Derivatives:'''  <math>\frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\tan x)=\sec^2 x</math>
|-
| 
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using the Chain Rule, we have
|-
| 
::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{\bigg(\frac{x^2-1}{x^2+1}\bigg)}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\
&&\\
& = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we need to calculate  <math>\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).</math>
|-
|To do this, we use the Quotient Rule. So, we have
|-
| 
::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\
&&\\
& = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}\bigg)}\\
&&\\
& = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{4x}{(x^2+1)^2}\bigg)}\\
&&\\
& = & \displaystyle{\frac{4x}{(x^2-1)(x^2+1)}}\\
&&\\
& = & \displaystyle{\frac{4x}{x^4-1}.}\\
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Again, we need to use the Chain Rule. We have
|-
|
::<math>g'(x)\,=\,8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We need to calculate  <math>\frac{d}{dx}\sqrt{1+x^3}.</math>
|-
|We use the Chain Rule again to get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg)}\\
&&\\
& = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\frac{1}{2}(1+x^3)^{-\frac{1}{2}}3x^2}\\
&&\\
& = & \displaystyle{8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -14px">f'(x)=\frac{4x}{x^4-1}</math>
|-
|'''(b)''' <math style="vertical-align: -18px">g'(x)=8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Find the derivatives of the following functions.
-<span class="exam">a) <math style="vertical-align: -16px">f(x)=\ln \bigg(\frac{x^2-1}{x^2+1}\bigg)</math>
+<span class="exam">(a) &nbsp; <math style="vertical-align: -14px">f(x)=\ln \bigg(\frac{x^2-1}{x^2+1}\bigg)</math>
-<span class="exam">b) <math style="vertical-align: -5px">g(x)=2\sin (4x)+4\tan (\sqrt{1+x^3})</math>
+<span class="exam">(b) &nbsp; <math style="vertical-align: -3px">g(x)=2\sin (4x)+4\tan (\sqrt{1+x^3})</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|For functions <math style="vertical-align: -5px">f(x)</math>&thinsp; and <math style="vertical-align: -5px">g(x)</math>, recall
+|'''1.''' '''Chain Rule'''
 |-
-|&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
 |-
+|'''2.''' '''Quotient Rule'''
 |-
-|&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp; <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.''' '''Trig Derivatives'''
 |-
-|&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\tan x)=\sec^2 x</math>
 |}
 '''Solution:'''
@@ Line 37: / Line 31: @@
 |Using the Chain Rule, we have
 |-
-|<br>
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{\bigg(\frac{x^2-1}{x^2+1}\bigg)}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\
 &&\\
@@ Line 48: / Line 42: @@
 !Step 2: &nbsp;
 |-
-|Now, we need to calculate &thinsp;<math>\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).</math>
+|Now, we need to calculate &nbsp;<math>\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).</math>
 |-
 |To do this, we use the Quotient Rule. So, we have
 |-
-|<br>
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\
 &&\\
@@ Line 71: / Line 65: @@
 !Step 1: &nbsp;
 |-
-|Again, we need to use the Chain Rule. We have
+|We need to use the Chain Rule. We have
 |-
+|
-::<math>g'(x)\,=\,8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg).</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>g'(x)\,=\,8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg).</math>
 |}
@@ Line 80: / Line 74: @@
 !Step 2: &nbsp;
 |-
-|We need to calculate&thinsp; <math>\frac{d}{dx}\sqrt{1+x^3}.</math>
+|We need to calculate &nbsp; <math>\frac{d}{dx}\sqrt{1+x^3}.</math>
 |-
 |We use the Chain Rule again to get
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{g'(x)} & = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg)}\\
 &&\\
@@ Line 93: / Line 87: @@
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' <math style="vertical-align: -14px">f'(x)=\frac{4x}{x^4-1}</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -14px">f'(x)=\frac{4x}{x^4-1}</math>
 |-
-|&nbsp;&nbsp; '''(b)''' <math style="vertical-align: -18px">g'(x)=8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -18px">g'(x)=8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 12: / Line 12: @@
 |&nbsp;
 |-
 |'''Chain Rule:'''&nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
 |-
 |&nbsp;
 |-
 |'''Quotient Rule:'''&nbsp; <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
 |-
 |&nbsp;
 |-
 |'''Trig Derivatives:'''&nbsp; <math>\frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\tan x)=\sec^2 x</math>
 |-
 |&nbsp;

009A Sample Final 1, Problem 3 - Revision history

MathAdmin at 16:08, 10 April 2017

MathAdmin at 22:13, 18 April 2016

MathAdmin at 19:03, 18 April 2016

MathAdmin: Created page with "Find the derivatives of the following functions. a) f(x)=\ln \bigg(\frac{x^2-1}{x^2+1}\bigg)..."

MathAdmin: Created page with "Find the derivatives of the following functions. a) $f(x)=\ln \bigg(\frac{x^2-1}{x^2+1}\bigg)$ ..."