Question

If $y=ta{n}^{-1}x$, find $\frac{d^{2}y}{d{x}^2}$
in terms of 𝑦 alone.

Answer 1

Differentiate w.r.t x to find $\frac{dy}{dx}$
Given : $y={\tan }^{-1}x$
Differentiating both sides w.r.t, x
$\Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2}$
Find the value of $\frac{d^{2}y}{d{x}^2}$
Again, differentiating both sides w.r.t. x
$\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{1}{1+x^2}\right)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-1{(1+x^2)}^{-2}.\frac{d}{dx}(1+x^2)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-2x}{{(1+x^2)}^2}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-2\tan y}{{(1+{\tan }^{2}y)}^2}$
$[\because y={\tan }^{-1}x\Rightarrow \tan y=x]$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{2tany}{secy}=-\frac{2sinyco{s}^{4}y}{cosy}$
$[\because \tan x=\frac{sinx}{cosx},secx=\frac{1}{cosx}]$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-2sinyco{s}^{3}y$
$Hence,\frac{d^{2}y}{d{x}^2}=-2sinyco{s}^{3}y$