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

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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
 
|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
 
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|-
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|<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
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|<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
 
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|<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>&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>
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::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
 
::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
 
|-
 
|-
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|we then have <math style="vertical-align: -21%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
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|we then have&thinsp; <math style="vertical-align: -21%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
 
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|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|-
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|We can now apply all three advanced techniques.  For example, to find the derivative <math>f'(x)</math>,
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|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
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|-
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|<br>
 +
::<math>\begin{array}{rcl}
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f'(x) & = & \displaystyle{\frac{\left((x+5)(x-1)\right)'x-(x+5)(x-1)(x)'}{x^{2}}}\\
 +
\\
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& = &  \displaystyle{\frac{\left[x^{2}+4x-5\right]'x-(x^{2}+4x-5)(x)'}{x^{2}}}\\
 +
\\
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& = &  \displaystyle{\frac{(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}}\\
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\\
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& = &  \displaystyle{\frac{2x^{2}-5x-x^{2}-4x+5}{x^{2}}}\\
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\\
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& = &  \displaystyle{\frac{x^{2}-9x+5}{x^{2}}}.
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\end{array}</math>
 
|
 
|
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Part (c): &nbsp;
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!Step 3: &nbsp;
 
|-
 
|-
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|We can choose to expand the second term, finding
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|We can now use the chain rule to find<br>
 
|-
 
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>e(x^{2}+2)^{2}=ex^{4}+4ex^{2}+4e.</math>
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|
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::<math>\begin{array}{rcl}
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y' & = & \left(g\circ f\right)'(x)\\
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\\
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& = & g'\left(f(x)\right)\cdot f'(x)\\
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\\& = & \displaystyle{\frac{x}{(x+5)(x-1)}\cdot\frac{x^{2}-9x+5}{x^{2}}}\\
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\\
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& = & \displaystyle{\frac{x^{2}-9x+5}{x^{3}+4x^{2}-5x}.}
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\end{array}</math>
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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>  
 
|-
 
|-
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|We then only require the product rule on the first term, so
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|
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::<math>y'=\displaystyle {\frac{x}{(x+5)(x-1)}\cdot\frac{(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}.}</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Final Answer: &nbsp;
 
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|-
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|&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>
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|<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>''']]
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

Revision as of 14:47, 14 May 2015

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

Foundations:  
This problem requires several advanced rules of differentiation. In particular, you need
The Chain Rule: If 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 f} and 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 g} are differentiable functions, then

     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 (f\circ g)'(x) = f'(g(x))\cdot g'(x).}

The Product Rule: If 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 f} and 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 g} are differentiable functions, then

     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 (fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).}

The Quotient Rule: If 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 f} and 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 g} are differentiable functions and 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 g(x) \neq 0}  , then

     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 \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 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 g(x)\,=\,\ln x} , and
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 f(x)\,=\,\frac{(x+5)(x-1)}{x},}
we then have  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\,=\,g\circ f(x)\,=\,g\left(f(x)\right).}
Step 2:  
We can now apply all three advanced techniques. For 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 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}.}}

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