009A Sample Midterm 3, Problem 6 - Revision history

Kayla Murray at 22:57, 18 February 2017

2017-02-18T22:57:15Z

2017-02-18T21:46:09Z

2017-02-18T21:45:11Z

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 <span class="exam"> Find the derivatives of the following functions. Do not simplify.
-<span class="exam">(a)&nbsp; <math>f(x)=\sin\bigg(\frac{x^{-3}}{e^{-x}}\bigg)</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{\frac{x^2+2}{x^2+4}}</math>
+<span class="exam">(b)&nbsp; <math style="vertical-align: -18px">g(x)=\sqrt{\frac{x^2+2}{x^2+4}}</math>
-<span class="exam">(c)&nbsp; <math>h(x)=(x+\cos^2x)^8</math>
+<span class="exam">(c)&nbsp; <math style="vertical-align: -6px">h(x)=(x+\cos^2x)^8</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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 <span class="exam">(a)&nbsp; <math>f(x)=\sin\bigg(\frac{x^{-3}}{e^{-x}}\bigg)</math>
-<span class="exam">b)&nbsp; <math>g(x)=\sqrt{\frac{x^2+2}{x^2+4}}</math>
+<span class="exam">(b)&nbsp; <math>g(x)=\sqrt{\frac{x^2+2}{x^2+4}}</math>
-<span class="exam">c)&nbsp; <math>h(x)=(x+\cos^2x)^8</math>
+<span class="exam">(c)&nbsp; <math>h(x)=(x+\cos^2x)^8</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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-<span class="exam">Find the derivatives of the following functions. Do not simplify.
+<span class="exam"> Find the derivatives of the following functions. Do not simplify.
-::<span class="exam">a)<math>f(x)=\sin\bigg(\frac{x^{-3}}{e^{-x}}\bigg)</math>
+<span class="exam">(a)&nbsp; <math>f(x)=\sin\bigg(\frac{x^{-3}}{e^{-x}}\bigg)</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Chain Rule
+|'''1.''' '''Chain Rule'''
 |-
-|'''2.''' Quotient Rule
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
 |}

@@ Line 21: / Line 21: @@
 !Step 1: &nbsp;
 |-
+|First, using the Chain Rule, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{x^{-3}}{e^{-x}}\bigg)'.</math>
 |}
@@ Line 29: / Line 29: @@
 !Step 2: &nbsp;
 |-
+|Now, using the Quotient Rule and Chain Rule, we have
 |-
+|
 |}
@@ Line 90: / Line 99: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-1)}{(e^{-x})^2}\bigg)</math>
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(2x)-(x^2+2)(2x)}{(x^2+4)^2}\bigg)</math>

@@ Line 38: / Line 38: @@
 !Step 1: &nbsp;
 |-
+|First, using the Chain Rule, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>g'(x)=\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{x^2+2}{x^2+4}\bigg)'.</math>
 |}
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 !Step 2: &nbsp;
 |-
+|Now, using the Quotient Rule, we have
 |-
+|
 |-
+|
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 |'''(a)'''
 |-
-|'''(b)'''
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(2x)-(x^2+2)(2x)}{(x^2+4)^2}\bigg)</math>
 |-
 |&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>8(x+\cos^2(x))^7(1-2\cos(x)\sin(x))</math>
 |}
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

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 ::
 |}
 '''Solution:'''