Question

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

Answer 1

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