If y= x +1 x , then dy dx at x = 1 is - Vidyadip

MCQ

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

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

✓

Answer

Correct option: D.

$0$

$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
$=\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}$
Differentiate both the sides with respect to $x$, we get
$\frac{1}{2}\text{x}^{\frac{1}{2}-1}+\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}$
$\frac{1}{2}\text{x}^{-\frac{1}{2}}-\Big(\frac{1}{2}\Big)\text{x}^{-\frac{3}{2}}$
Putting $x = 1,$ we get
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=1}=\frac{1}{2}\times1-\frac{1}{2}\times1=0$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is $0.$

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