022 Exam 2 Sample A, Problem 1 - Revision history

MathAdmin at 05:42, 19 January 2017

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MathAdmin at 05:40, 19 January 2017

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MathAdmin at 17:22, 16 May 2015

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MathAdmin at 14:43, 16 May 2015

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MathAdmin at 22:52, 14 May 2015

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MathAdmin at 22:47, 14 May 2015

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MathAdmin at 22:47, 14 May 2015

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MathAdmin at 22:16, 14 May 2015

2015-05-14T22:16:34Z

MathAdmin: Created page with "Find the derivative of $y\,=\,\ln \frac{(x+5)(x-1)}{x}.$ {| class="mw-collapsible mw-collapsed" style = "te..."

2015-05-14T22:16:01Z

Created page with "Find the derivative of <math style="vertical-align: -42%">y\,=\,\ln \frac{(x+5)(x-1)}{x}.</math> {| class="mw-collapsible mw-collapsed" style = "te..."

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Find the derivative of  <math style="vertical-align: -42%">y\,=\,\ln \frac{(x+5)(x-1)}{x}.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|This problem requires several advanced rules of differentiation. In particular, you need
|-
|'''The Chain Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
|-

|      <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
|-
| '''The Product Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
|-
|      <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
|-
| '''The Quotient Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math> , then
|-
|      <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
|-
| 
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We need to identify the composed functions in order to apply the chain rule. Note that if we set <math style="vertical-align: -25%">g(x)\,=\,\ln x</math>, and
|-
|
::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
|-
|we then have <math style="vertical-align: -25%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We can now apply all three advanced techniques. For example, to find the derivative <math>f'(x)</math>,
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Part (c):  
|-
|We can choose to expand the second term, finding
|-
|          <math>e(x^{2}+2)^{2}=ex^{4}+4ex^{2}+4e.</math>
|-
|We then only require the product rule on the first term, so
|-
|          <math>h'(x)\,=\,(4x)'\cdot\sin(x)+4x\cdot(\sin(x))'+\left(ex^{4}+4ex^{2}+4e\right)'\,=\,4\sin(x)+4x\cos(x)+4ex^{3}+8ex.</math>
|}

[[022_Exam_2_Sample_A|'''Return to Sample Exam''']]

@@ Line 10: / Line 10: @@
 |and
 |-
-|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln \left( \frac{x}{y}\right) = \ln x - \ln y,</math>
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln \left( \frac{x}{y}\right) = \ln x - \ln y,</math>
 |-
 |You will also need to apply
@@ Line 18: / Line 17: @@
 |-
-|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 |-
@@ Line 43: / Line 42: @@
 !Step 2: &nbsp;
 |-
-|We can differentiate term-by-term, applying the chain rule to the first two to find
+|We can differentiate term-by-term, applying the chain rule to the first two terms to find
 |-
 |<br>
@@ Line 58: / Line 57: @@
 !Final Answer: &nbsp;
 |-
-|<math>y'\,=\,\displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}.</math>
+|<br><math>y'\,=\,\displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}.</math>
 |}
 [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 4: / Line 4: @@
 !Foundations: &nbsp;
 |-
-|This problem requires several advanced rules of differentiation.  In particular, you need
+|This problem is best approached through properties of logarithms.  Remember that
 |-
-|'''The Chain Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
+|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln (xy) = \ln x + \ln y,</math>
 |-
+|and
 |-
-|<br>'''The Product Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
+|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln \left( \frac{x}{y}\right) = \ln x - \ln y,</math>
 |-
 |-
-|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math>&thinsp;, then
+|You will also need to apply
 |-
-|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
+|'''The Chain Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
 |-
-|Additionally, we will need our power rule for differentiation:
 |-
-−
-::<math style="vertical-align: -21%;">\left(x^n\right)'\,=\,nx^{n-1},</math> for <math style="vertical-align: -25%;">n\neq 0</math>,
+|Finally, recall that the derivative of natural log is
 |-
+|
@@ Line 36: / Line 33: @@
 !Step 1: &nbsp;
 |-
-|We need to identify the composed functions in order to apply the chain rule.  Note that if we set <math style="vertical-align: -21%">g(x)\,=\,\ln x</math>, and
+|We can use the log rules to rewrite our function as
 |-
+|
-::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
+::<math>y\,=\,\ln (x+5)+\ln(x-1)-\ln(x).</math>
 |-
 |}
@@ Line 47: / Line 43: @@
 !Step 2: &nbsp;
 |-
-|We can now apply the advanced techniques.  For <math style="vertical-align: -20%">f'(x)</math>, we can avoid using the product rule by first multiplying out the denominator.  Then by the quotient rule,
+|We can differentiate term-by-term, applying the chain rule to the first two to find
 |-
 |<br>
 ::<math>\begin{array}{rcl}
-f'(x) & = & \displaystyle{\frac{\left[(x+5)(x-1)\right]'x-(x+5)(x-1)(x)'}{x^{2}}}\\
+y' & = & \displaystyle{\frac{1}{x+5}\cdot(x+5)'+\frac{1}{x-1}\cdot(x-1)'+\frac{1}{x}}\\
 \\
-  & = &  \displaystyle{\frac{\left(x^{2}+4x-5\right)'x-(x^{2}+4x-5)(x)'}{x^{2}}}\\
+  & = &  \displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}.
 \end{array}</math>
+|
 |}
@@ Line 88: / Line 58: @@
 !Final Answer: &nbsp;
 |-
-|<math>y'\,=\,\displaystyle{\frac{x^{2}+5}{x^{3}+4x^{2}-5x}.}</math>
+|<math>y'\,=\,\displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}.</math>
 |}
 [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 47: / Line 47: @@
 !Step 2: &nbsp;
 |-
-|We can now apply all three advanced techniques.  For <math style="vertical-align: -20%">f'(x)</math>, we must use both the quotient and product rule to find
+|We can now apply the advanced techniques.  For <math style="vertical-align: -20%">f'(x)</math>, we can avoid using the product rule by first multiplying out the denominator.  Then by the quotient rule,
 |-
 |<br>
 ::<math>\begin{array}{rcl}
-f'(x) & = & \displaystyle{\frac{\left((x+5)(x-1)\right)'x-(x+5)(x-1)(x)'}{x^{2}}}\\
+f'(x) & = & \displaystyle{\frac{\left[(x+5)(x-1)\right]'x-(x+5)(x-1)(x)'}{x^{2}}}\\
 \\
-  & = &  \displaystyle{\frac{\left[x^{2}+4x-5\right]'x-(x^{2}+4x-5)(x)'}{x^{2}}}\\
+  & = &  \displaystyle{\frac{\left(x^{2}+4x-5\right)'x-(x^{2}+4x-5)(x)'}{x^{2}}}\\
 \\
   & = &  \displaystyle{\frac{(2x+4)x-(x^{2}+4x-5)(1)}{x^{2}}}\\

@@ Line 55: / Line 55: @@
   & = &  \displaystyle{\frac{\left[x^{2}+4x-5\right]'x-(x^{2}+4x-5)(x)'}{x^{2}}}\\
 \\
-  & = &  \displaystyle{\frac{(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}}\\
+  & = &  \displaystyle{\frac{(2x+4)x-(x^{2}+4x-5)(1)}{x^{2}}}\\
 \\
-& = &  \displaystyle{\frac{2x^{2}-5x-x^{2}-4x+5}{x^{2}}}\\
+& = &  \displaystyle{\frac{2x^{2}+4x-x^{2}-4x+5}{x^{2}}}\\
   \\
-& = &  \displaystyle{\frac{x^{2}-9x+5}{x^{2}}}.
+& = &  \displaystyle{\frac{x^{2}+5}{x^{2}}}.
 \end{array}</math>
+|
@@ Line 74: / Line 74: @@
 \\
 & = & g'\left(f(x)\right)\cdot f'(x)\\
-\\& = & \displaystyle{\frac{x}{(x+5)(x-1)}\cdot\frac{x^{2}-9x+5}{x^{2}}}\\
+\\& = & \displaystyle{\frac{x}{(x+5)(x-1)}\cdot\frac{x^{2}+5}{x^{2}}}\\
 \\
-& = & \displaystyle{\frac{x^{2}-9x+5}{x^{3}+4x^{2}-5x}.}
+& = & \displaystyle{\frac{x^{2}+5}{x^{3}+4x^{2}-5x}.}
 \end{array}</math>
 Note that many teachers <u>'''do not'''</u> prefer a cleaned up answer, and may request that you <u>'''do not simplify'''</u>.  In this case, we could write the answer as<br>
 |-
+|
-::<math>y'=\displaystyle {\frac{x}{(x+5)(x-1)}\cdot\frac{(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}.}</math>
+::<math>y'=\displaystyle {\frac{x}{(x+5)(x-1)}\cdot\frac{(2x+4)x-(x^{2}+4x-5)(1)}{x^{2}}.}</math>
 |}
@@ Line 88: / Line 88: @@
 !Final Answer: &nbsp;
 |-
-|<math>y'\,=\,\displaystyle{\frac{x^{2}-9x+5}{x^{3}+4x^{2}-5x}.}</math>
+|<math>y'\,=\,\displaystyle{\frac{x^{2}+5}{x^{3}+4x^{2}-5x}.}</math>
 |}
 [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 19: / Line 19: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
 |-
 |<br>
 |}

@@ Line 15: / Line 15: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
 |-
-|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -20%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math>&thinsp;, then
+|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math>&thinsp;, then
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>

@@ Line 15: / Line 15: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
 |-
-|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math>&thinsp;, then
+|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -20%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math>&thinsp;, then
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
@@ Line 32: / Line 32: @@
 ::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
 |-
-|we then have <math style="vertical-align: -21%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
+|we then have&thinsp; <math style="vertical-align: -21%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
 |}
@@ Line 38: / Line 38: @@
 !Step 2: &nbsp;
 |-
-|We can now apply all three advanced techniques.  For example, to find the derivative <math>f'(x)</math>,
+|We can now apply all three advanced techniques.  For <math style="vertical-align: -20%">f'(x)</math>, we must use both the quotient and product rule to find
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Part (c): &nbsp;
+!Step 3: &nbsp;
 |-
-|We can choose to expand the second term, finding
+|We can now use the chain rule to find<br>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>e(x^{2}+2)^{2}=ex^{4}+4ex^{2}+4e.</math>
 |-
-|We then only require the product rule on the first term, so
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>h'(x)\,=\,(4x)'\cdot\sin(x)+4x\cdot(\sin(x))'+\left(ex^{4}+4ex^{2}+4e\right)'\,=\,4\sin(x)+4x\cos(x)+4ex^{3}+8ex.</math>
+|<math>y'\,=\,\displaystyle{\frac{x^{2}-9x+5}{x^{3}+4x^{2}-5x}.}</math>
 |}
 [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 27: / Line 27: @@
 !Step 1: &nbsp;
 |-
-|We need to identify the composed functions in order to apply the chain rule.  Note that if we set <math style="vertical-align: -25%">g(x)\,=\,\ln x</math>, and
+|We need to identify the composed functions in order to apply the chain rule.  Note that if we set <math style="vertical-align: -20%">g(x)\,=\,\ln x</math>, and
 |-
+|
 ::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
 |-
-|we then have <math style="vertical-align: -25%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
+|we then have <math style="vertical-align: -22%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
 |}

022 Exam 2 Sample A, Problem 1 - Revision history

MathAdmin at 05:42, 19 January 2017

MathAdmin at 05:40, 19 January 2017

MathAdmin at 17:22, 16 May 2015

MathAdmin at 14:43, 16 May 2015

MathAdmin at 22:52, 14 May 2015

MathAdmin at 22:47, 14 May 2015

MathAdmin at 22:47, 14 May 2015

MathAdmin at 22:17, 14 May 2015

MathAdmin at 22:16, 14 May 2015

MathAdmin: Created page with "Find the derivative of y\,=\,\ln \frac{(x+5)(x-1)}{x}. {| class="mw-collapsible mw-collapsed" style = "te..."

MathAdmin: Created page with "Find the derivative of $y\,=\,\ln \frac{(x+5)(x-1)}{x}.$ {| class="mw-collapsible mw-collapsed" style = "te..."