Question

Differentiate

$\frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},\frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}} \\ \mathrm{By}\mathrm{rationalising}\mathrm{we}\mathrm{get}, \\ \frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}\times \frac{\sqrt{{\mathrm{x}}^2+1}+\sqrt{{\mathrm{x}}^2-1}}{\sqrt{{\mathrm{x}}^2+1}+\sqrt{{\mathrm{x}}^2-1}} \\ =\frac{{\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)}^2}{{\left(\sqrt{x^2+1}\right)}^2-{\left(\sqrt{x^2-1}\right)}^2} \\ =\frac{{\left(\sqrt{x^2+1}\right)}^2+{\left(\sqrt{x^2-1}\right)}^2+2\left(\sqrt{x^2+1}\right)\left(\sqrt{x^2-1}\right)}{x^2+1-x^2+1} \\ =\frac{x^2+1+x^2-1+2\sqrt{x^4-1}}{2} \\ =\frac{2x^2+2\sqrt{x^4-1}}{2} \\ =x^2+\sqrt{x^4-1} \\ \mathrm{Now},\mathrm{Let}y=x^2+\sqrt{x^4-1} \end{gathered}$$

Differentiate it with respect to x we get,

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(x^2+\sqrt{x^4-1}\right) \\ =2x+\frac{1}{2\sqrt{x^4-1}}\times \frac{d}{dx}\left(x^4-1\right) \\ =2x+\frac{1}{2\sqrt{x^4-1}}\times \left(4x^3\right) \\ =2x+\frac{2x^3}{\sqrt{x^4-1}} \end{gathered}$$