If y=x+1x , then dydx at x=1 is

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

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

Solution:

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

On differentiating w.r.t. x we get $\frac{dy}{dx} = \frac{1}{2 \sqrt{x}} - \frac{1}{2 x^{3 / 2}}$

At, $x = 1, \frac{dy}{dx} = \frac{1}{2 \sqrt{1}} - \frac{1}{2 (1)^{3 / 2}} = \frac{1}{2} - \frac{1}{2} = 0$

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