Question

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

Answer 1

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