Question

Differentiate, ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to ${\tan }^{-1}\left(x\right)$

Answer 1

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

Differentiate on both sides w.r.t x

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

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

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

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

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

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

Let $t={\tan }^{-1}\left(x\right)$

Differentiate on both sides w.r.t x

$\frac{dt}{dx}=\frac{1}{1+x^2}$

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

$=\frac{\frac{dy}{dx}}{\frac{dt}{dx}}$

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

$\therefore \frac{dy}{dx}=\frac{1}{2}$