Difference between revisions of "022 Exam 2 Sample B, Problem 3"
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(Created page with "<span class="exam"> Find the derivative: <math style="vertical-align: -18%">f(x) \,=\, 2x^3e^{3x+5}</math>.") |
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| − | <span class="exam"> Find the derivative | + | <span class="exam"> Find the derivative of <math style="vertical-align: -18%">f(x) \,=\, 2x^3e^{3x+5}</math>. |
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| + | {| 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 | ||
| + | |- | ||
| + | |||
| + | |<br> <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math> | ||
| + | |- | ||
| + | |<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> <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</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 the exponential function, <math style="vertical-align: 5%">e^x</math>: | ||
| + | |- | ||
| + | | | ||
| + | ::<math>(e^x)'\,=\,e^x.</math> | ||
| + | |<br> | ||
| + | |} | ||
| + | |||
| + | '''Solution:''' | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 1: | ||
| + | |- | ||
| + | |We need to start by identifying the two functions that are being multiplied together so we can apply the product rule. Let's call <math style="vertical-align: -20%">g(x)\,=\,2x^3,\,</math> and <math style="vertical-align: -20%">\,h(x) \, = \, e^{3x + 5}</math>, so <math style="vertical-align: -20%">f(x)=g(x)\cdot h(x)</math>. | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 2: | ||
| + | |- | ||
| + | |We can now apply the advanced techniques.This allows us to see that | ||
| + | |- | ||
| + | | | ||
| + | ::<math>\begin{array}{rcl} | ||
| + | f'(x)&=&2(x^3)' e^{3x+5}+2x^3(e^{3x+5})' \\ | ||
| + | &=&6x^2e^{3x+5}+2x^3(3e^{3x+5})\\ | ||
| + | & = &6x^2e^{3x+5}+6x^3e^{3x+5}. | ||
| + | \end{array}</math> | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Final Answer: | ||
| + | |- | ||
| + | | | ||
| + | ::<math>f'(x)\,=\,6x^2e^{3x+5}+6x^3e^{3x+5}. | ||
| + | </math> | ||
| + | |} | ||
| + | |||
| + | [[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']] | ||
Latest revision as of 07:50, 17 May 2015
Find the derivative of .
| 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).} | |
| Additionally, we will need our power rule for differentiation: | |
| |
| as well as the derivative of the exponential function, 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 e^x} : | |
|
Solution:
| Step 1: |
|---|
| We need to start by identifying the two functions that are being multiplied together so we can apply the product rule. Let's call 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)\,=\,2x^3,\,} 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 \,h(x) \, = \, e^{3x + 5}} , so 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)=g(x)\cdot h(x)} . |
| Step 2: |
|---|
| We can now apply the advanced techniques.This allows us to see that |
|
| Final Answer: |
|---|
|