Question

Differentiate ${\tan }^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$ with respect to $x.$

Answer 1

$\frac{d\left[ta{n}^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]\right]}{dx}$

$=\frac{1}{1+{\left[\frac{\sqrt{1+x^2}-1}{x}\right]}^2}\times \frac{d\left[\frac{\sqrt{1+x^2}-1}{x}\right]}{dx}$

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

$=\frac{x^2}{2(1+x^2-\sqrt{1+x^2})}\times \frac{x\times \frac{x}{\sqrt{1+x^2}}-\sqrt{1+x^2}+1}{x^2}$

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

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

$=\frac{1}{2(1+x^2)}$