|
|
| (One intermediate revision by the same user not shown) |
| Line 1: |
Line 1: |
| − | <span class="exam"> Let <math style="vertical-align: -3px">y=\sqrt{2x+5},x\ge 0.</math> | + | <span class="exam"> Let <math style="vertical-align: -3px">y=3\sqrt{2x+5},x\ge 0.</math> |
| | | | |
| | <span class="exam">(a) Use the definition of the derivative to compute <math style="vertical-align: -13px">\frac{dy}{dx}.</math> | | <span class="exam">(a) Use the definition of the derivative to compute <math style="vertical-align: -13px">\frac{dy}{dx}.</math> |
| Line 12: |
Line 12: |
| | | | |
| | | | |
| − | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Foundations:
| |
| − | |-
| |
| − | |Recall
| |
| − | |-
| |
| − | | <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | '''Solution:'''
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |Let <math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
| |
| − | |-
| |
| − | |Using the limit definition of the derivative, we have
| |
| − | |-
| |
| − | |
| |
| − | <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\sqrt{-2(x+h)+5}-3\sqrt{-2x+5}}{h}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3\sqrt{-2x+-2h+5}-3\sqrt{-2x+5}}{h}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{\sqrt{-2x+-2h+5}-\sqrt{-2x+5}}{h}.}
| |
| − | \end{array}</math>
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Now, we multiply the numerator and denominator by the conjugate of the numerator.
| |
| − | |-
| |
| − | |Hence, we have
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \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})}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{(-2x+-2h+5)-(-2x+5)}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2h}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2}{\sqrt{-2x+-2h+5}+\sqrt{-2x+5}}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{3\frac{-2}{\sqrt{-2x+5}+\sqrt{-2x+5}}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-\frac{3}{\sqrt{-2x+5}}.}
| |
| − | \end{array}</math>
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | <math>-\frac{3}{\sqrt{-2x+5}}</math>
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| | [[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 
at 
Solution
Detailed Solution
Return to Sample Exam