Question

Derivative of $f\left(x\right)=4\sqrt{x}+\frac{1}{\sqrt{x}}$ is

• A. $\frac{2}{\sqrt{x}}+\frac{1}{2x\sqrt{x}}$
• B. $\frac{2}{\sqrt{x}}-\frac{1}{2x\sqrt{x}}$
• C. $\frac{1}{2x\sqrt{x}}-\frac{2}{\sqrt{x}}$
• D. $-\frac{1}{2x\sqrt{x}}-\frac{2}{\sqrt{x}}$

Answer 1

The correct option is B $\frac{2}{\sqrt{x}}-\frac{1}{2x\sqrt{x}}$

Given, $f\left(x\right)=4\sqrt{x}+\frac{1}{\sqrt{x}}$
So, derivative will be
$\frac{df\left(x\right)}{dx}=\frac{d(4\sqrt{x}+\frac{1}{\sqrt{x}})}{dx}=\frac{d(4x^{\frac{1}{2}}+x^{\frac{-1}{2}})}{dx}$
$\Rightarrow 4\frac{d{x}^{\frac{1}{2}}}{dx}+\frac{d{x}^{\frac{-1}{2}}}{dx}=4\times \frac{1}{2}\left(x^{\frac{-1}{2}}\right)-\frac{1}{2}\left(x^{\frac{-3}{2}}\right)$
$\Rightarrow \frac{2}{\sqrt{x}}-\frac{1}{2x\sqrt{x}}$