Question

If $f\left(x\right)=\frac{x-4}{2\sqrt{x}}$, then $f^{\prime }\left(1\right)$ is

• A. $\frac{5}{4}$
• B. $\frac{4}{5}$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

We have, $f\left(x\right)=\frac{x-4}{2\sqrt{x}}$
$\therefore f^{\prime }\left(x\right)=\frac{1}{2}\left\{\frac{d}{dx}\left(\frac{x-4}{\sqrt{x}}\right)\right\}$
$=\frac{1}{2}\frac{\left\{\sqrt{x}\frac{d}{dx}\left(x-4\right)-\left(x-4\right)\frac{d}{dx}\sqrt{x}\right\}}{{\left(\sqrt{x}\right)}^2}$
$=\frac{1}{2}\left\{\left(\sqrt{x}\times 1\right)-\left(x-4\right)\times \frac{1}{2\sqrt{x}}\right\}\times \frac{1}{x}$
$=\frac{1}{2}\left(\sqrt{x}-\frac{x-4}{2\sqrt{x}}\right)\frac{1}{x}$
$\therefore f^{\prime }\left(1\right)=\frac{1}{2}\left(1-\frac{\left(-3\right)}{2}\right)\times 1=\frac{5}{4}$