Question

If $y=\sqrt{x}+\frac{1}{\sqrt{x}}$, show that $2x\frac{dy}{dx}+y=2\sqrt{x}.$

Answer 1

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

Differentiating with respect to x, we get

$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}+\left(\frac{-1}{2\sqrt[x]{xx}}\right)$

$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}.$

$\frac{dy}{dx}=\frac{x-1}{2x\sqrt{x}}$

$2x\frac{dy}{dx}=\frac{x-1}{\sqrt{x}}$

$2x\frac{dy}{dx}=\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}}.$

$2x\frac{dy}{dx}=\sqrt{x}-\frac{1}{\sqrt{x}}$

$2x\frac{dy}{dx}=\sqrt{x}-\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}$

$2x\frac{dy}{dx}=2\sqrt{x}-y$

$\Rightarrow 2x\frac{dy}{dx}+y=2\sqrt{x}$