Second Derivatives

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

put $x = tan\theta$

$y = 2tan^{-1}(\frac{\sqrt{1+ tan^2\theta} -1 }{tan\theta})$

$y = 2tan^{-1}(\frac{sec\theta -1}{tan\theta})$

Since $sec\theta = \frac{1}{cos\theta}$

$y = 2tan^{-1}(\frac{1 - cos\theta}{tan\theta cos\theta})$

$y = 2tan^{-1}(\frac{2sin^2\frac{\theta}{2}}{2sin\frac{\theta}{2} cos\frac{\theta}{2}})$

$y = 2tan^{-1}(tan\frac{\theta}{2})$

$y = 2\frac{\theta}{2} => \theta$

$y = tan^{-1}x$

Differentiating with respect to x.

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

Using the quotient rule for second derivatives

$\frac{dy^2}{dx^2} = \frac{0(1+x^2) - (2x).1}{(1+x^2)^2}$

$\frac{d^2y}{dx^2} = \frac{-2\times2}{(1+2^2)^2}$

$(\frac{d^2y}{dx^2})_(x=2) = \boxed{\frac{-4}{25}}$

$\begin{aligned} y & = 2 \tan^{-1} \left(\frac {\sqrt{1+x^2}-1}x\right) \\ \tan \frac y2 & = \frac {\sqrt{1+x^2}-1}x & \small \color{#3D99F6} \text{Let } x = \tan \theta \\ & = \frac {\sec \theta - 1}{\tan \theta} & \small \color{#3D99F6} \text{Multiply up and down by } \cos \theta \\ & = \frac {1 - \cos \theta}{\sin \theta} & \small \color{#3D99F6} \text{Let } t = \tan \frac \theta 2 \\ & = \frac {1 - \frac {1-t^2}{1+t^2}}{\frac {2t}{1+t^2}} \\ & = \frac {1+t^2-1+t^2}{2t} \\ & = t = \tan \frac \theta 2 \end{aligned}$

$\begin{aligned} \implies y & = \theta \\ \tan y & = x & \small \color{#3D99F6} \text{Differentiate both sides w.r.t } x \\ \sec^2 y \cdot \frac {dy}{dx} & = 1 \\ \frac {dy}{dx} & = \cos^2 y \\ \frac {d^2y}{dx^2} & = - 2 \cos y \sin y \cdot \frac {dy}{dx} \\ & = - 2 \cos^3 y \sin y \\ \frac {d^2y}{dx^2} \bigg|_{x=2} & = - 2 \left(\frac 1{\sqrt 5}\right)^3 \left(\frac 2{\sqrt 5}\right) & \small \color{#3D99F6} \tan y = x = 2 \\ & = \boxed{-\dfrac 4{25}} & \small \color{#3D99F6} \implies \sin y = \frac 2{\sqrt 5}, \ \cos y = \frac 1{\sqrt 5} \end{aligned}$