If, \(y=\sqrt x+\frac{1}{\sqrt x}\) prove that :- \(2x.\frac{dy}{

1.	If, \(y=\sqrt x+\frac{1}{\sqrt x}\) prove that :- \(2x.\frac{dy}{dx}+y=2\sqrt x.\)
Answer» Given, \(y=\sqrt x+\frac{1}{\sqrt{x}}\) ⇒ \(\frac{dy}{dx}=\frac{d}{dx}(\sqrt x+\frac{1}{\sqrt x})\) \(=\frac{1}{2}x^\frac{1}{2}+(-\frac{1}{2})x^\frac{-3}{2}\) \(=\frac{1}{2}.\frac{1}{\sqrt{x}}-\frac{1}{2x\sqrt{x}}\) ∴ L.H.S = \(2x\frac{dy}{dx}+y\) \(=2x(\frac{1}{2x\sqrt x}-\frac{1}{2x\sqrt x})+\sqrt{x}+\frac{1}{\sqrt{x}}\) \(=\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}\) \(=2\sqrt{x}\) = R.H.S

If, \(y=\sqrt x+\frac{1}{\sqrt x}\) prove that :- \(2x.\frac{dy}{dx}+y=2\sqrt x.\)

Answer»

Given,

\(y=\sqrt x+\frac{1}{\sqrt{x}}\)

⇒ \(\frac{dy}{dx}=\frac{d}{dx}(\sqrt x+\frac{1}{\sqrt x})\)

\(=\frac{1}{2}x^\frac{1}{2}+(-\frac{1}{2})x^\frac{-3}{2}\)

\(=\frac{1}{2}.\frac{1}{\sqrt{x}}-\frac{1}{2x\sqrt{x}}\)

∴ L.H.S

= \(2x\frac{dy}{dx}+y\)

\(=2x(\frac{1}{2x\sqrt x}-\frac{1}{2x\sqrt x})+\sqrt{x}+\frac{1}{\sqrt{x}}\)

\(=\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}\)

\(=2\sqrt{x}\)

= R.H.S