009A Sample Midterm 3, Problem 5 - Revision history

Kayla Murray at 01:58, 7 November 2017

2017-11-07T01:58:39Z

2017-11-03T17:49:39Z

2017-11-03T17:47:59Z

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@@ Line 9: / Line 9: @@
 <hr>
-(insert picture of handwritten solution)
+[[009A Sample Midterm 3, Problem 5 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
-[[009A Sample Midterm 3, Problem 5 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
+[[File:9ASM3P5.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the derivatives of the following functions. Do not simplify.
-<span class="exam">(a)&nbsp; <math style="vertical-align: -16px">f(x)=\frac{(3x-5)(-x^{-2}+4x)}{x^{\frac{4}{5}}}</math>
+<span class="exam">(a)&nbsp; <math style="vertical-align: -16px">f(x)=\sin\bigg(\frac{x^{-3}}{e^{-x}}\bigg)</math>
-<span class="exam">(b)&nbsp; <math>g(x)=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{\pi}</math>&nbsp; for &nbsp;<math style="vertical-align: 0px">x>0.</math>
+<span class="exam">(b)&nbsp; <math style="vertical-align: -18px">g(x)=\sqrt{\frac{x^2+2}{x^2+4}}</math>
 <hr>

@@ Line 5: / Line 5: @@
 <span class="exam">(b)&nbsp; <math>g(x)=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{\pi}</math>&nbsp; for &nbsp;<math style="vertical-align: 0px">x>0.</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-'''Solution:'''
+[[009A Sample Midterm 3, Problem 5 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

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 \displaystyle{f'(x)} & = & \displaystyle{\frac{x^{\frac{4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac{4}{5}})'}{(x^{\frac{4}{5}})^2}}\\
 &&\\
-& = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(-x^{-2}+4x)'+(3x-5)'(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}}\\
+& = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(-x^{-2}+4x)'+(3x-5)'(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{-\frac{1}{5}})}{(x^{\frac{4}{5}})^2}}\\
 &&\\
-& = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}.}
+& = & \displaystyle{\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{-\frac{1}{5}})}{(x^{\frac{4}{5}})^2}.}
 \end{array}</math>
 |}
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 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{-\frac{1}{5}})}{(x^{\frac{4}{5}})^2}</math>
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}</math>
 |}
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

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 \displaystyle{g'(x)} & = & \displaystyle{(\sqrt{x})'+\bigg(\frac{1}{\sqrt{x}}\bigg)'+(\sqrt{\pi})'}\\
 &&\\
-& = & \displaystyle{\frac{1}{2}x^{\frac{-1}{2}}+\frac{-1}{2}x^{\frac{-3}{2}}+0}\\
+& = & \displaystyle{\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}+0}\\
 &&\\
-& = & \displaystyle{\frac{1}{2}x^{\frac{-1}{2}}+\frac{-1}{2}x^{\frac{-3}{2}}.}
+& = & \displaystyle{\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}.}
 \end{array}</math>
 |}
@@ Line 83: / Line 83: @@
 |&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{x^{\frac{4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)(\frac{4}{5}x^{\frac{-1}{5}})}{(x^{\frac{4}{5}})^2}</math>
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{1}{2}x^{\frac{-1}{2}}+\frac{-1}{2}x^{\frac{-3}{2}}</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}</math>
 |}
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

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 !Step 2: &nbsp;
 |-
-|Since <math style="vertical-align: 0px">\pi</math> is a constant, <math style="vertical-align: -3px">\sqrt{\pi}</math> is also a constant.
+|Since &nbsp;<math style="vertical-align: 0px">\pi</math>&nbsp; is a constant, &nbsp;<math style="vertical-align: -3px">\sqrt{\pi}</math>&nbsp; is also a constant.
 |-
 |Hence,