#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 59 maths

Hint: you must know the rule of solving derivation of functions

Given: $\mathrm{y}=\sqrt{x+1}+\sqrt{x-1}$

Prove : $\sqrt{x^{2}-1} \frac{d y}{d x}=\frac{1}{2} y$

Solution:

Let  $\mathrm{y}=\sqrt{x+1}+\sqrt{x-1}$

Differentiate with respect to x

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

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

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

∴ Proved

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