Question

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

Answer 1

$\mathrm{Here},y=\log \left(\sqrt{x-1}-\sqrt{x+1}\right)$
Differentiate it with respect to x we get,
$$\begin{gathered} \frac{\mathit{d}y}{\mathit{d}x}=\frac{d}{dx}\log \left(\sqrt{x-1}-\sqrt{x+1}\right) \\ =\frac{1}{\left(\sqrt{x-1}-\sqrt{x+1}\right)}\frac{d}{dx}\left(\sqrt{x-1}-\sqrt{x+1}\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right] \\ =\frac{1}{\left(\sqrt{x-1}-\sqrt{x+1}\right)}\left[\frac{d}{dx}\sqrt{x-1}-\frac{d}{dx}\sqrt{x+1}\right] \\ =\frac{1}{\left(\sqrt{x-1}-\sqrt{x+1}\right)}\left[\frac{1}{2}{\left(x-1\right)}^{\frac{-1}{2}}-\frac{1}{2}{\left(x+1\right)}^{\frac{-1}{2}}\right] \\ =\frac{1}{2}\frac{1}{\left(\sqrt{x-1}-\sqrt{x+1}\right)}\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right) \\ =\frac{1}{2}\frac{1}{\left(\sqrt{x-1}-\sqrt{x+1}\right)}\left\{\frac{-\left(\sqrt{x-1}-\sqrt{x+1}\right)}{\left(\sqrt{x-1}\right)\left(\sqrt{x+1}\right)}\right\} \\ =\frac{-1}{2}\left(\frac{1}{\left(\sqrt{x-1}\right)\left(\sqrt{x+1}\right)}\right) \\ =\frac{-1}{2\sqrt{x^2-1}} \\ \mathrm{So},\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{-1}{2\sqrt{x^2-1}} \end{gathered}$$