If y=x+1 x , prove that 2 x d y d x =x-1 x - Vidyadip

Question

If $\mathrm{y}=\sqrt{x}+\frac{1}{\sqrt{x}}$, prove that $2 \mathrm{x} \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}}$

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Answer

We have, $\mathrm{y}=\sqrt{x}+\frac{1}{\sqrt{x}}$
Differentiate with respect to x ,
$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$
$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(\sqrt{x})+\frac{d}{d x}\left(\frac{1}{\sqrt{x}}\right)$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{x}}+\left(\frac{-\frac{1}{2 \sqrt{x}}}{x}\right)$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{x}}-\frac{1}{2 x \sqrt{x}}$
$\Rightarrow \frac{d y}{d x}=\frac{x-1}{2 x \sqrt{x}}$
$\Rightarrow 2 x \frac{d y}{d x}=\frac{x-1}{\sqrt{x}}$
$\Rightarrow 2 x \frac{d y}{d x}=\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}}$
$\Rightarrow 2 x \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}}$

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