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

Question

Quizplus · Accepted Answer

Using the definition of the derivative, we have:
$$g'(x) = \lim_{h 	o 0} \frac{g(x+h)-g(x)}{h}$$
$$= \lim_{h 	o 0} \frac{\sqrt{2-3(x+h)} - \sqrt{2-3x}}{h}$$
Rationalizing the denominator,
$$= \lim_{h 	o 0} \frac{\sqrt{2-3(x+h)} - \sqrt{2-3x}}{h} \cdot \frac{\sqrt{2-3(x+h)} + \sqrt{2-3x}}{\sqrt{2-3(x+h)} + \sqrt{2-3x}}$$
$$= \lim_{h 	o 0} \frac{(2-3(x+h))-(2-3x)}{h(\sqrt{2-3(x+h)} + \sqrt{2-3x})}$$
$$= \lim_{h 	o 0} \frac{-3h}{h(\sqrt{2-3(x+h)} + \sqrt{2-3x})}$$
$$= \lim_{h 	o 0} \frac{-3}{\sqrt{2-3(x+h)} + \sqrt{2-3x}}$$
$$= -\frac{3}{2\sqrt{2-3x}} \cdot \frac{2\sqrt{2-3x}}{2\sqrt{2-3x}+2\sqrt{2-3x}}$$
$$= -\frac{3}{2\sqrt{2-3x}(2\sqrt{2-3x}+2\sqrt{2-3x})}$$
$$= -\frac{3}{2(2-3x)}\cdot\frac{1}{\sqrt{2-3x}}$$
$$= -\frac{3}{2(2-3x)\sqrt{2-3x}}$$
$$= -\frac{3}{2}\cdot\frac{1}{\sqrt{2-3x}\cdot(2-3x)}$$
Therefore, the correct answer is D.

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

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