Question

If $y=\sqrt{x}+\frac{1}{\sqrt{x}}$, then $\frac{dy}{dx}$ at $x=1$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $1$
• C. $\frac{1}{2}$
• D. $0$

Answer 1

The correct option is D $0$

Given $y=\sqrt{x}+\frac{1}{\sqrt{x}}$,

To find $\frac{dy}{dx}$ at $x=1$

$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2x^{\frac{3}{2}}}$ $(\frac{d{x}^n}{dx}=n{x}^{n-1})$

at $x=1,\frac{dy}{dx}=\frac{1}{2}-\frac{1}{2}$
$\frac{dy}{dx}=0\text{at}x=1$
Hence, the correct option is (d)