If $g ( x ) = \sqrt { 2 - 3 x }$, use the definition of derivative to find $g ^ { \prime } ( x )$
A) $g ^ { \prime } ( x ) = - \frac { 1 } { 2 } ( 2 - 3 x ) ^ { - 1 / 2 }$
B) $g ^ { \prime } ( x ) = - ( 2 - 3 x ) ^ { - 1 / 2 }$
C)
$g ^ { \prime } ( x ) = - \frac { 3 } { 2 } ( 2 - 3 x ) ^ { 1 / 2 }$
D)
$g ^ { \prime } ( x ) = - \frac { 3 } { 2 } ( 2 - 3 x ) ^ { - 1 / 2 }$
E) None of these

Question

Quizplus · Accepted Answer

Using the definition of derivative, we have:
$$ g ^ { \prime } ( x ) = \lim_{h 	o 0} \frac{g(x+h)-g(x)}{h} $$
Substituting the given function, we get:
\begin{align*}
g ^ { \prime } ( x ) &= \lim_{h 	o 0} \frac{\sqrt{2-3(x+h)}-\sqrt{2-3x}}{h} \
&= \lim_{h 	o 0} \frac{\sqrt{2-3x-3h}-\sqrt{2-3x}}{h} \
&= \lim_{h 	o 0} \frac{(2-3x-3h)-(2-3x)}{h(\sqrt{2-3x-3h}+\sqrt{2-3x})} \
&= \lim_{h 	o 0} \frac{-3h}{h(\sqrt{2-3x-3h}+\sqrt{2-3x})}\
&= \lim_{h 	o 0} \frac{-3}{\sqrt{2-3x-3h}+\sqrt{2-3x}}\
&= -\frac{3}{2\sqrt{2-3x}}
\end{align*}
Therefore, the correct choice is D.

If g(x)=2−3xg ( x ) = \sqrt { 2 - 3 x }g(x)=2−3x​

If $g ( x ) = \sqrt { 2 - 3 x }$