022 Exam 2 Sample B, Problem 1 - Revision history

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MathAdmin: Created page with " Problem 1 ]] == Find the derivative of $y\,=\,\ln \frac{(x+1)..."$

2015-05-15T23:53:17Z

Created page with " Problem 1 ]] == Find the derivative of <math style="vertical-align: -60%">y\,=\,\ln \frac{(x+1)..."

New page

Problem 1 ]] ==
Find the derivative of  <math style="vertical-align: -60%">y\,=\,\ln \frac{(x+1)^4}{(2x - 5)(x + 4)}.</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>
|-
|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>,
|-
|as well as the derivative of natural log:
|-
|
::<math>\left(\ln x\right)'\,=\,\frac{1}{x}.</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: -21%">g(x)\,=\,\ln x</math>, and
|-
|
::<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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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
|-
| 
::<math>\begin{array}{rcl}
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{(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}}\\
\\
& = & \displaystyle{\frac{2x^{2}-5x-x^{2}-4x+5}{x^{2}}}\\
\\
& = & \displaystyle{\frac{x^{2}-9x+5}{x^{2}}}.
\end{array}</math>
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|We can now use the chain rule to find 
|-
|
::<math>\begin{array}{rcl}
y' & = & \left(g\circ f\right)'(x)\\
\\
& = & 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^{2}-9x+5}{x^{3}+4x^{2}-5x}.}
\end{array}</math>
Note that many teachers '''do not''' prefer a cleaned up answer, and may request that you '''do not simplify'''. In this case, we could write the answer as 
|-
|
::<math>y'=\displaystyle {\frac{x}{(x+5)(x-1)}\cdot\frac{(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}.}</math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|<math>y'\,=\,\displaystyle{\frac{x^{2}-9x+5}{x^{3}+4x^{2}-5x}.}</math>
|}

[[022_Exam_2_Sample_B|'''Return to Sample Exam''']]

@@ Line 40: / Line 40: @@
 |<br>
 ::<math>\begin{array}{rcl}
-y' & = & \displaystyle{\ln \frac{(x+1)^4}{(2x - 5)(x + 4)}}\\
+y & = & \displaystyle{\ln \frac{(x+1)^4}{(2x - 5)(x + 4)}}\\
 \\
   & = &  4\ln (x+1)-\ln(2x-5)-\ln (x+4).
@@ Line 47: / Line 47: @@
+|
 |-
 |-
 |}

@@ 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>
 |-
+|while
 |-
-|<br>'''The Product Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln \left( \frac{x}{y}\right) = \ln x - \ln y,</math>
 |-
-|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
+|and
 |-
-|<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
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln \left( x^n \right) = n\ln x,</math>
 |-
-|<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>
+|You will also need to apply
 |-
-|Additionally, we will need our power rule for differentiation:
+|'''The Chain Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
 |-
-−
-::<math style="vertical-align: -21%;">\left(x^n\right)'\,=\,nx^{n-1},</math> for <math style="vertical-align: -25%;">n\neq 0</math>,
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 |-
-|as well as the derivative of natural log:
 |-
+|
@@ Line 36: / Line 36: @@
 !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
 |-
+|
 |-
-|we then have&thinsp; <math style="vertical-align: -21%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
+|-
 |}
@@ Line 47: / Line 53: @@
 !Step 2: &nbsp;
 |-
-|We can now apply all three advanced techniques.  For <math style="vertical-align: -20%">f'(x)</math>, we can use both the quotient and product rule to find
+|We can differentiate term-by-term, applying the chain rule to each term to find
 |-
 |<br>
 ::<math>\begin{array}{rcl}
-f'(x)&=&\displaystyle{\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2}} \\
+y' & = & \displaystyle{4 \cdot \frac{1}{x+1}\cdot(x+1)' - \frac{1}{2x-5}\cdot(2x-5)' - \frac{1}{x+4} \cdot (x+4)'}\\
 \\
-&=&\displaystyle{\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}.}
+ & = &  \displaystyle{\frac{4}{x+1}-\frac{2}{2x-5}-\frac{1}{x+4}}.
 \end{array}</math>
+|
 |}
@@ Line 78: / Line 68: @@
 !Final Answer: &nbsp;
 |-
-|<math>y'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. </math>
+|<br><math>y'\,=\,\displaystyle{\frac{4}{x+1}-\frac{2}{2x-5}-\frac{1}{x+4}}.</math>
 |}
 [[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']]

@@ Line 72: / Line 72: @@
 &=&\displaystyle{\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. }
 \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>
+Note that many teachers <u>'''do not'''</u> prefer a cleaned up answer, and may request that you <u>'''do not simplify'''</u>.  This problem seems like it would be of that type, as it doesn't simplify too well.  Nevertheless, it's always a good idea to ask the teacher if you aren't sure of his or her intent.
 |}

@@ 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 all three advanced techniques.  For <math style="vertical-align: -20%">f'(x)</math>, we can use both the quotient and product rule to find
 |-
 |<br>
 ::<math>\begin{array}{rcl}
-f'(x)&=&\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2} \\
+f'(x)&=&\displaystyle{\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2}} \\
-&=&\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}.
+\\
 \end{array}</math>
+|
@@ Line 67: / Line 68: @@
 \\
 & = & g'\left(f(x)\right)\cdot f'(x)\\
-\\& = &\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2} \\
+\\& = &\displaystyle{\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2}} \\
 \\
-&=&\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}
+&=&\displaystyle{\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. }
 \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'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2} </math>
+::<math>y'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. </math>
 |}
@@ Line 81: / Line 82: @@
 !Final Answer: &nbsp;
 |-
-|<math>y'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2} </math>
+|<math>y'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. </math>
 |}
 [[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']]

@@ Line 39: / Line 39: @@
 |-
+|
-::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
+::<math>f(x)\,=\,\frac{(x+1)^4}{(2x - 5)(x + 4)},</math>
 |-
 |we then have&thinsp; <math style="vertical-align: -21%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
@@ Line 51: / Line 51: @@
 |<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)&=&\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2} \\
-\\
+&=&\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}.
 \end{array}</math>
+|
@@ Line 74: / Line 67: @@
 \\
 & = & g'\left(f(x)\right)\cdot f'(x)\\
-\\& = & \displaystyle{\frac{x}{(x+5)(x-1)}\cdot\frac{x^{2}-9x+5}{x^{2}}}\\
+\\& = &\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2} \\
 \\
-& = & \displaystyle{\frac{x^{2}-9x+5}{x^{3}+4x^{2}-5x}.}
+&=&\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}
 \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'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2} </math>
 |}
@@ Line 88: / Line 81: @@
 !Final Answer: &nbsp;
 |-
-|<math>y'\,=\,\displaystyle{\frac{x^{2}-9x+5}{x^{3}+4x^{2}-5x}.}</math>
+|<math>y'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2} </math>
 |}
 [[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 23:54, 15 May 2015
Line 1:		Line 1:
−	~~<span style="font-size:80%"> Problem 1 </span></span>]] ==~~
	<span class="exam">Find the derivative of  <math style="vertical-align: -60%">y\,=\,\ln \frac{(x+1)^4}{(2x - 5)(x + 4)}.</math>		<span class="exam">Find the derivative of  <math style="vertical-align: -60%">y\,=\,\ln \frac{(x+1)^4}{(2x - 5)(x + 4)}.</math>