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

Basic idea:

$\sqrt{S e c o n d}$ order derivative is nothing but derivative of derivative i.e. $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$

$\sqrt{T h e}$ 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:

$\frac{d f}{d x}=\frac{d v}{d t} \times \frac{d t}{d x}$

$\sqrt{P r o d u c t}$ 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, $y=x^{3}+\tan x$

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

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

So lets first find $\frac{d y}{d x}$ and differentiate it again.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}+\tan \mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}\right)+\frac{\mathrm{d}}{\mathrm{dx}}(\tan \mathrm{x})$

$\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\tan \mathrm{x})=\sec ^{2} \times \& \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}\right]$

$=3 \mathrm{x}^{2}+\mathrm{sec}^{2} \mathrm{x}$

$\therefore \frac{d y}{d x}=3 x^{2}+\sec ^{2} x$

Differentiating again with respect to $x$ :

$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(3 x^{2}+\sec ^{2} x\right)=\frac{d}{d x}\left(3 x^{2}\right)+\frac{d}{d x}\left(\sec ^{2} x\right)$

$\frac{d^{2} y}{d x^{2}}=6 x+2 \sec x \sec x \tan x$

[ differentiated $\sec ^{2} \times$ using chain rule, let $t=\sec \times a n d z=t^{2} \therefore \frac{d z}{d x}=\frac{d z}{d t} \times \frac{d t}{d x}$ ]

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=6 \mathrm{x}+2 \sec ^{2} \mathrm{x} \tan \mathrm{x}$