022 Exam 2 Sample B, Problem 3 - Revision history

MathAdmin at 14:50, 17 May 2015

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MathAdmin at 14:46, 17 May 2015

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MathAdmin at 00:41, 16 May 2015

2015-05-16T00:41:16Z

MathAdmin at 00:38, 16 May 2015

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MathAdmin at 00:36, 16 May 2015

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MathAdmin at 00:35, 16 May 2015

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MathAdmin at 00:25, 16 May 2015

2015-05-16T00:25:55Z

MathAdmin: Created page with " Find the derivative: $f(x) \,=\, 2x^3e^{3x+5}$ ."

2015-05-16T00:22:07Z

Created page with "<span class="exam"> Find the derivative: <math style="vertical-align: -18%">f(x) \,=\, 2x^3e^{3x+5}</math>."

New page

<span class="exam"> Find the derivative: <math style="vertical-align: -18%">f(x) \,=\, 2x^3e^{3x+5}</math>.

@@ Line 33: / Line 33: @@
 !Step 1: &nbsp;
 |-
-|We need to start by identifying the two functions that are being multiplied together so we can apply the product rule.
+|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>.
 |}
@@ Line 42: / Line 39: @@
 !Step 2: &nbsp;
 |-
-|We can now apply the three advanced techniques.This allows us to see that
+|We can now apply the advanced techniques.This allows us to see that
 |-
+|
-<math>\begin{array}{rcl}
+::<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}
+& = &6x^2e^{3x+5}+6x^3e^{3x+5}.
 \end{array}</math>
 |}
@@ Line 56: / Line 53: @@
 |-
+|
-<math>6x^2e^{3x+5}+6x^3e^{3x+5}
+::<math>f'(x)\,=\,6x^2e^{3x+5}+6x^3e^{3x+5}.
 </math>
 |}
 [[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']]

@@ Line 21: / Line 21: @@
 ::<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>e^x</math>:
+|as well as the derivative of the exponential function, <math style="vertical-align: 5%">e^x</math>:
 |-
+|
-::<math>\left(e^{f(x)}\right)'\,=\,f'(x)\cdot e^{f(x)}.</math>
+::<math>(e^x)'\,=\,e^x.</math>
 |<br>
 |}

← Older revision		Revision as of 00:41, 16 May 2015
Line 59:		Line 59:
	</math>		</math>
	\|}		\|}
		+
		+	[[022_Exam_2_Sample_B\|'''<u>Return to Sample Exam</u>''']]

@@ Line 36: / Line 36: @@
 |-
+|
-::<math>g(x)\,=\,2x^3,</math> and <math>h(x) \, = \, e^{3x + 5}</math>
+::<math>g(x)\,=\,2x^3,\,</math> and <math>\,h(x) \, = \, e^{3x + 5}</math>
 |}

@@ Line 36: / Line 36: @@
 |-
+|
-::<math>g(x)\,=\,2x^3,</math>
+::<math>g(x)\,=\,2x^3,</math> and <math>h(x) \, = \, e^{3x + 5}</math>
 |}
@@ Line 49: / Line 47: @@
 <math>\begin{array}{rcl}
 f'(x)&=&2(x^3)' e^{3x+5}+2x^3(e^{3x+5})' \\
-&=&6x^2e^{3x+5}+6x^3e^{3x+5}
+&=&6x^2e^{3x+5}+2x^3(3e^{3x+5})\\
 \end{array}</math>
 |}

@@ Line 24: / Line 24: @@
 |-
+|
-::<math>\left(e^{f(x)}\right)'\,=\,\left(f(x)\right)'\cdot e^{f(x)}.</math>
+::<math>\left(e^{f(x)}\right)'\,=\,f'(x)\cdot e^{f(x)}.</math>
 |<br>
 |}
@@ Line 33: / 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 need to start by identifying the two functions that are being multiplied together so we can apply the product rule.
 |-
+|
-::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
+::<math>g(x)\,=\,2x^3,</math>
 |-
-|we then have&thinsp; <math style="vertical-align: -21%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
+|and <math>h(x) \, = \, e^{3x + 5}</math>
 |}

← Older revision		Revision as of 00:25, 16 May 2015
Line 1:		Line 1:
−	<span class="exam"> Find the derivative: <math style="vertical-align: -18%">f(x) \,=\, 2x^3e^{3x+5}</math>.	+	<span class="exam"> Find the derivative of <math style="vertical-align: -18%">f(x) \,=\, 2x^3e^{3x+5}</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
		+	\|-
		+
		+	\|<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>e^x</math>:
		+	\|-
		+	\|
		+	::<math>\left(e^{f(x)}\right)'\,=\,\left(f(x)\right)'\cdot e^{f(x)}.</math>
		+	\|<br>
		+	\|}
		+
		+	'''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>
		+	\|}