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
$$
g ^ { \prime } ( x ) = \lim _ { h ightarrow 0 } \frac { g ( x + h ) - g ( x ) } { h }=\lim _ { h ightarrow 0 } \frac { \sqrt { 2 - 3 ( x + h ) } - \sqrt { 2 - 3 x } } { h }
$$
We can simplify this expression by multiplying the numerator and denominator of the fraction by the conjugate of the numerator:
\begin{align*}
g ^ { \prime } ( x ) &=\lim _ { h ightarrow 0 } \frac { \sqrt { 2 - 3 ( x + h ) } - \sqrt { 2 - 3 x } } { h } \cdot \frac {\sqrt { 2 - 3 ( x + h ) } + \sqrt { 2 - 3 x }}{\sqrt { 2 - 3 ( x + h ) } + \sqrt { 2 - 3 x }} \
&=\lim _ { h ightarrow 0 } \frac { 2 - 3 ( x + h ) - ( 2 - 3 x ) } { h ( \sqrt { 2 - 3 ( x + h ) } + \sqrt { 2 - 3 x } ) } \
&=\lim _ { h ightarrow 0 } \frac { -3 h } { h ( \sqrt { 2 - 3 ( x + h ) } + \sqrt { 2 - 3 x } ) } \
&=\frac { -3 } { 2 \sqrt { 2 - 3 x } }
\end{align*}
Therefore, the answer is D.

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

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