Difference between revisions of "022 Exam 2 Sample A, Problem 1"

Revision as of 15:47, 14 May 2015

Find the derivative of $y\,=\,\ln {\frac {(x+5)(x-1)}{x}}.$

Foundations:
This problem requires several advanced rules of differentiation. In particular, you need
The Chain Rule: If $f$ and $g$ are differentiable functions, then
$(f\circ g)'(x)=f'(g(x))\cdot g'(x).$
The Product Rule: If $f$ and $g$ are differentiable functions, then
$(fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).$
The Quotient Rule: If $f$ and $g$ are differentiable functions and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)\neq 0} , then
$\left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.$

Solution:

Step 1:
We need to identify the composed functions in order to apply the chain rule. Note that if we set $g(x)\,=\,\ln x$ , and
$f(x)\,=\,{\frac {(x+5)(x-1)}{x}},$
we then have $y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).$

Step 2:

We can now apply all three advanced techniques. For

f'(x)

, we must use both the quotient and product rule to find

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Step 3:

We can now use the chain rule to find

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'=\displaystyle {\frac{x}{(x+5)(x-1)}\cdot\frac{(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}.}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'\,=\,\displaystyle{\frac{x^{2}-9x+5}{x^{3}+4x^{2}-5x}.}}

Return to Sample Exam

@@ 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
+|-
+|<br>
+::<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;"
-!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>
+|
+::<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 <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>
 |-
-|We then only require the product rule on the first term, so
+|
+::<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: &nbsp;
 |-
-|&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>''']]

Difference between revisions of "022 Exam 2 Sample A, Problem 1"

Revision as of 15:47, 14 May 2015

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