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| | <span class="exam">Use the definition of the derivative to find <math>\frac{dy}{dx}</math> for the function <math style="vertical-align: -12px">y=\frac{1+x}{3x}.</math> | | <span class="exam">Use the definition of the derivative to find <math>\frac{dy}{dx}</math> for the function <math style="vertical-align: -12px">y=\frac{1+x}{3x}.</math> |
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| | + | (insert picture of handwritten solution) |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
| − | !Foundations:
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| − | |Recall
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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:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |Let <math style="vertical-align: -14px">f(x)=\frac{1+x}{3x}.</math>
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| − | |Using the limit definition of derivative, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(\frac{1+(x+h)}{3(x+h)})-(\frac{1+x}{3x})}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(\frac{1+x+h}{3x+3h})-(\frac{1+x}{h})}{h}.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, we get a common denominator for the fractions in the numerator.
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| − | |Hence, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0}\frac{\frac{(1+x+h)3x}{(3x+3h)(3x)}-\frac{(1+x)(3x+3h)}{(3x+3h)(3x)}}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0}\frac{\frac{3x+3x^2+3xh-(3x+3h+3x^2+3hx)}{(3x+3h)(3x)}}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{-3h}{h(3x+3h)(3x)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{-3}{(3x+3h)(3x)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-3}{(3x)(3x)}}\\
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| − | & = & \displaystyle{-\frac{1}{3x^2}.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math>-\frac{1}{3x^2}</math>
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| − | |}
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| | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
