#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 63

Hint: you must know the rule of solving derivation of functions.

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

Prove    $2 x \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}}$

Solution:

Let  $y=\sqrt{x}+\frac{1}{\sqrt{x}}$

Differentiate with respect to x,

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left\{\sqrt{x}+\frac{1}{\sqrt{x}}\right\} \\\\ &\frac{d y}{d x}=\frac{d}{d x}(\sqrt{x})+\frac{d}{d x}\left(\frac{1}{\sqrt{x}}\right) \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}+\left(\frac{-1}{2 x \sqrt{x}}\right) \\\\ &\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}-\frac{1}{2 x \sqrt{x}} \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{x-1}{2 x \sqrt{x}} \\\\ &2 x \frac{d y}{d x}=\frac{x-1}{\sqrt{x}} \end{aligned}

\begin{aligned} &2 x \frac{d y}{d x}=\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}} \\\\ &2 x \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}} \end{aligned}

$\therefore$ Proved