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 the derivative, we have:

$$g^\prime(x) = \lim_{h 	o 0} \frac{g(x+h) - g(x)}{h}$$

Substituting in the expression for $g(x)$, we have:

\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{2-3(x+h)} - \sqrt{2-3x}}{h} \cdot \frac{\sqrt{2-3(x+h)} + \sqrt{2-3x}}{\sqrt{2-3(x+h)} + \sqrt{2-3x}} \
&= \lim_{h 	o 0} \frac{(2-3(x+h)) - (2-3x)}{h(\sqrt{2-3(x+h)} + \sqrt{2-3x})} \
&= \lim_{h 	o 0} \frac{-3h}{h(\sqrt{2-3(x+h)} + \sqrt{2-3x})} \
&= \lim_{h 	o 0} \frac{-3}{\sqrt{2-3(x+h)} + \sqrt{2-3x}} \
&= -\frac{3}{\sqrt{2-3x} + \sqrt{2-3x}} \
&= -\frac{3}{2\sqrt{2-3x}} \
&= -\frac{3}{2}\cdot (2-3x)^{-1/2}
\end{align*}

Therefore, the correct answer is D.

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

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