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. -1
  

Answer 1

The correct option is A $\frac{5}{4}$


$f\left(x\right)=\frac{x-4}{2\sqrt{x}}$

$=\frac{1}{2}\sqrt{x}-\frac{2}{\sqrt{x}}$

$=\frac{1}{2}{x}^{\frac{1}{2}}-2x^{-\frac{1}{2}}$

Differentiating both sides with respect to $x$, we get

$f^{\prime }\left(x\right)=\frac{1}{2}\times \frac{1}{2}{x}^{\frac{1}{2}-1}-2\times (-\frac{1}{2})x^{-\frac{1}{2}-1}$ $[\because f(x)=x^n\Rightarrow f^{\prime }(x)=n{x}^{n-1}]$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{4}{x}^{-\frac{1}{2}}+x^{-\frac{3}{2}}$

$=\frac{1}{4\sqrt{x}}+\frac{1}{x\sqrt{x}}$

$=\frac{x+4}{4x\sqrt{x}}$

$\therefore f^{\prime }\left(1\right)=\frac{1+4}{4\times 1\sqrt{1}}=\frac{1+4}{4}=\frac{5}{4}$

Hence, the correct answer is option A.