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:

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

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

$\sqrt{P}$ Product rule of differentiation- $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{uv})=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}$

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

Let's solve now:

Given, $y=\log (\log x)$

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

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

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\log \log \mathrm{x})$

Let $y=\log t$ and $t=\log x$

Using chain rule of differentiation:

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{t}} \times \frac{1}{\mathrm{x}}=\frac{1}{\mathrm{x} \log \mathrm{x}}\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}\right]$

Again differentiating w.r.t $\mathrm{x}$ :

As, $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{u} \times \mathrm{v}$

Where $u=\frac{1}{x}$ and $v=\frac{1}{\log x}$

$\therefore$ using product rule of differentiation:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}$

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{1}{x} \frac{d}{d x}\left(\frac{1}{\log x}\right)+\frac{1}{\log x} \frac{d}{d x}\left(\frac{1}{x}\right)$ [ use chain rule to find $\frac{d}{d x}\left(\frac{1}{\log x}\right)$ ]

$\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}=-\frac{1}{\mathrm{x}^{2}(\log \mathrm{x})^{2}}-\frac{1}{\mathrm{x}^{2} \log \mathrm{x}}\left[\because \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}\right.$ and $\left.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}\right]$

$\therefore \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{1}{\mathrm{x}^{2}(\log \mathrm{x})^{2}}-\frac{1}{\mathrm{x}^{2} \log \mathrm{x}}$