#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 58

Hint: you must know the rule of solving derivative of logarithm functions

Given: $\log (\sqrt{x-1}-\sqrt{x+1})$

Solution:

Let  $y=\log (\sqrt{x-1}-\sqrt{x+1})$

Differentiate with respect to x

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x} \log (\sqrt{x-1}-\sqrt{x+1}) \\\\ &\frac{d y}{d x}=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})} \cdot \frac{d}{d x}(\sqrt{x-1}-\sqrt{x+1}) \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})} \cdot\left[\frac{d}{d x}(\sqrt{x-1})-\frac{d}{d x} \sqrt{x+1}\right] \\\\ &\frac{d y}{d x}=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})} \cdot\left[\frac{1}{2}(x-1)^{\frac{-1}{2}}-\frac{1}{2}(x+1)^{\frac{-1}{2}}\right] \end{aligned}

$\frac{d y}{d x}=\frac{1}{2} \frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right)$

\begin{aligned} &\frac{d y}{d x}=\frac{1}{2(\sqrt{x-1}-\sqrt{x+1})}\left(\frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x-1})(\sqrt{x+1})}\right) \\\\ &\frac{d y}{d x}=\frac{-1}{2 \sqrt{x^{2}-1}} \end{aligned}

∴ Proved