009A Sample Midterm 3, Problem 3

Use the definition of the derivative to compute   ${\frac {dy}{dx}}$   for  $y=3{\sqrt {-2x+5}}.$

Foundations:
Recall
$f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}$

Solution:

Step 1:
Let  $f(x)=3{\sqrt {-2x+5}}.$
Using the limit definition of the derivative, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3{\sqrt {-2(x+h)+5}}-3{\sqrt {-2x+5}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3{\sqrt {-2x+-2h+5}}-3{\sqrt {-2x+5}}}{h}}}\\&&\\&=&\displaystyle {3\lim _{h\rightarrow 0}{\frac {{\sqrt {-2x+-2h+5}}-{\sqrt {-2x+5}}}{h}}.}\end{array}}$

Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, we have
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 \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}}

Final Answer:
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 -\frac{3}{\sqrt{-2x+5}}}

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