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| − | <span class="exam"> Use the definition of the derivative to compute <math>\frac{dy}{dx}</math> for <math style="vertical-align: -4px">y=3\sqrt{-2x+5}.</math> | + | <span class="exam"> Let <math style="vertical-align: -3px">y=3\sqrt{2x+5},x\ge 0.</math> |
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| | + | <span class="exam">(a) Use the definition of the derivative to compute <math style="vertical-align: -13px">\frac{dy}{dx}.</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <span class="exam">(b) Find the equation of the tangent line to <math style="vertical-align: -3px">y=3\sqrt{2x+5}</math> at <math style="vertical-align: -3px">(2,9).</math> |
| − | !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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| − | |}
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| | + | [[009A Sample Midterm 3, Problem 3 Solution|'''<u>Solution</u>''']] |
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Midterm 3, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Step 1:
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| − | |Let <math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
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| − | |Using the limit definition of the 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{3\sqrt{-2(x+h)+5}-3\sqrt{-2x+5}}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3\sqrt{-2x+-2h+5}-3\sqrt{-2x+5}}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{\sqrt{-2x+-2h+5}-\sqrt{-2x+5}}{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 multiply the numerator and denominator by the conjugate of 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{3\lim_{h\rightarrow 0} \frac{(\sqrt{-2x+-2h+5}-\sqrt{-2x+5})}{h} \frac{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
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| − | &&\\
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| − | & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{(-2x+-2h+5)-(-2x+5)}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
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| − | &&\\
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| − | & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2h}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
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| − | &&\\
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| − | & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2}{\sqrt{-2x+-2h+5}+\sqrt{-2x+5}}}\\
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| − | &&\\
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| − | & = & \displaystyle{3\frac{-2}{\sqrt{-2x+5}+\sqrt{-2x+5}}}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{3}{\sqrt{-2x+5}}.}
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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{dy}{dx}=-\frac{3}{\sqrt{-2x+5}}</math>
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| − | |}
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| | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
Let 
(a) Use the definition of the derivative to compute 
(b) Find the equation of the tangent line to 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=3\sqrt{2x+5}}
at 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 (2,9).}
Solution
Detailed Solution
Return to Sample Exam