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

Solution:

$\sqrt{B a s i c}$ Idea: Second 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:

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

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

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let's solve now:

Given, $y=x^{3} \log x$

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

As $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

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

$\therefore \frac{d y}{d x}=\frac{d}{d x}\left(x^{3} \log x\right)$

Let $u=x^{3}$ and $v=\log x$

As, $y=u v$

$\therefore$ Using product rule of differentiation:

$\frac{d y}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}$

$\therefore \frac{d y}{d x}=x^{3} \frac{d}{d x}(\log x)+\log x \frac{d}{d x} x^{3}$'

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

$\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}\right.$ and $\left.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}\right]$

Again differentiating w.r.t $x$ :

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

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

Again using the product rule :

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

$\left[\because \frac{d}{d x}(\log x)=\frac{1}{x}\right.$ and $\left.\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$

$\frac{d^{2} y}{d x^{2}}=6 x \log x+\frac{3 x^{2}}{x}+2 x$

$\frac{d^{2} y}{d x^{2}}=6 x \log x+5 x$