Question

If $y=\log \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$, prove that $\frac{dy}{dx}=\frac{x-1}{2x\left(x+1\right)}$

Answer 1

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