$$\text { Find the derivative of the function } f ( x ) = \sqrt { 7 x - 3 } \text { using the limiting process. }$$
A) $f ^ { t } ( x ) = \frac { 7 } { 2 \sqrt { 7 x - 3 } }$
B) $f ^ { t } ( x ) = - \frac { 7 } { 2 \sqrt { 7 x - 3 } }$
C) $f ^ { t } ( x ) = - \frac { 7 x } { \sqrt { 7 x - 3 } }$
D) $f ^ {t } ( x ) = \frac { 7 } { 2 } \sqrt { 7 x - 3 }$
E) $f ^ { t } ( x ) = - \frac { 7 } { \sqrt { 7 x - 3 } }$

Question

$$\text { Find the derivative of the function } f ( x ) = \sqrt { 7 x - 3 } \text { using the limiting process. }$$ 
A) $f ^ { t } ( x ) = \frac { 7 } { 2 \sqrt { 7 x - 3 } }$
B) $f ^ { t } ( x ) = - \frac { 7 } { 2 \sqrt { 7 x - 3 } }$
C) $f ^ { t } ( x ) = - \frac { 7 x } { \sqrt { 7 x - 3 } }$
D) $f ^ {t } ( x ) = \frac { 7 } { 2 } \sqrt { 7 x - 3 }$
E) $f ^ { t } ( x ) = - \frac { 7 } { \sqrt { 7 x - 3 } }$

Quizplus · Accepted Answer

Let's start with the definition of the derivative:
$$f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
We have 
$$f(x) = \sqrt{7x-3}$$
Let's plug that into the definition of the derivative above and simplify the expression as much as possible.
\begin{align*}
f'(x) &= \lim_{h\to0}\frac{\sqrt{7(x+h)-3}-\sqrt{7x-3}}{h}\
&= \lim_{h\to0}\frac{\sqrt{7x+7h-3}-\sqrt{7x-3}}{h}\
&= \lim_{h\to0}\frac{\sqrt{7x+7h-3}-\sqrt{7x-3}}{h}\cdot\frac{\sqrt{7x+7h-3}+\sqrt{7x-3}}{\sqrt{7x+7h-3}+\sqrt{7x-3}}\
&= \lim_{h\to0}\frac{7x+7h-3-7x+3}{h(\sqrt{7x+7h-3}+\sqrt{7x-3})}\
&= \lim_{h\to0}\frac{7h}{h(\sqrt{7x+7h-3}+\sqrt{7x-3})}\
&= \lim_{h\to0}\frac{7}{\sqrt{7x+7h-3}+\sqrt{7x-3}}\
&= \frac{7}{2\sqrt{7x-3}} \text{ (Rationalizing the denominator)}
\end{align*}

Therefore, the correct answer is A.

Find the derivative of the function f(x)=7x−3 using the limiting process. \text { Find the derivative of the function } f ( x ) = \sqrt { 7 x - 3 } \text { using the limiting process. } Find the derivative of the function f(x)=7x−3​ using the limiting process.

$\text { Find the derivative of the function } f ( x ) = \sqrt { 7 x - 3 } \text { using the limiting process. }$