Question

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

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

Answer 1

The correct option is D $0$

We have,

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

On taking differentiating both sides w.r.t $x$, we get

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

Since, $x=1$

Therefore,

${\frac{dy}{dx}|}_{x=1}=\frac{1}{2\sqrt{1}}-\frac{1}{2{\left(1\right)}^{\frac{3}{2}}}$

${\frac{dy}{dx}|}_{x=1}=\frac{1}{2}-\frac{1}{2}$

${\frac{dy}{dx}|}_{x=1}=0$

Hence, this is the answer.