Difference between revisions of "009A Sample Final 3, Problem 3"
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!Foundations: | !Foundations: | ||
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− | | | + | |<math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math> |
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'''Solution:''' | '''Solution:''' | ||
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
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− | | | + | |Using the limit definition of derivative, we have |
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− | | | + | | <math>\begin{array}{rcl} |
+ | \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(3(x+h)-(x+h)^2)-(3x-x^2)}{h}}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3x+3h-(x^2+2xh+h^2)-3x+x^2}{h}}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3h-2xh-h^2}{h}.} | ||
+ | \end{array}</math> | ||
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!Step 2: | !Step 2: | ||
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− | | | + | |Now, we have |
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− | | | + | | <math>\begin{array}{rcl} |
− | + | \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3h-2xh-h^2}{h}}\\ | |
− | + | &&\\ | |
− | + | & = & \displaystyle{\lim_{h\rightarrow 0} 3-2x-h}\\ | |
− | + | &&\\ | |
− | + | & = & \displaystyle{3-2x.} | |
− | + | \end{array}</math> | |
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!Final Answer: | !Final Answer: | ||
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− | | | + | | <math>f'(x)=3-2x</math> |
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[[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
Revision as of 10:37, 6 March 2017
Find the derivative of the following function using the limit definition of the derivative:
Foundations: |
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Solution:
Step 1: |
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Using the limit definition of derivative, we have |
Step 2: |
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Now, we have |
Final Answer: |
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