Find the second order derivatives of each of the following functions:

Solution:

Basic idea:

$\sqrt{T h e}$ idea of chain rule of differentiation: If $\mathrm{f}$ is any real-valued function which is the composition of two functions $u$ and $v$, i.e. $f=v(u(x))$. For the sake of simplicity just assume $t=u(x)$

Then $f=v(t)$. By chain rule, we can write the derivative of $f$ w.r.t to $x$ as:

$\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}$

$\sqrt{P r o d u c t}$ rule of differentiation $-\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}$

Let's solve now:

Given, $y=\tan ^{-1} x$

We have to find $\frac{d^{2} y}{d x^{2}}$

As $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$

So lets first find $d y / d x$ and differentiate it again.

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

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

Differentiating again with respect to $\mathrm{x}$ :

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

Differentiating $\frac{1}{1+x^{2}}$ using chain rule,

let $\mathrm{t}=1+\mathrm{x}^{2}$ and $\mathrm{z}=1 / \mathrm{t}$

$\because \frac{\mathrm{d} z}{\mathrm{dx}}=\frac{\mathrm{dz}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}[$ from chain rule of differentiation $]$

$\therefore \frac{\mathrm{dz}}{\mathrm{dx}}=\frac{-1}{\mathrm{t}^{2}} \times 2 \mathrm{X}=-\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\left[\because \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}\right]$

$\therefore \frac{d^{2} y}{d x^{2}}=-\frac{2 x}{\left(1+x^{2}\right)^{2}}$