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

$\sqrt{T h e}$ 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- $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{uv})=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}$

Let's solve now:

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

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

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

So lets first find dy/dx and differentiate it again.

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

differentiating $\sin (\log x)$ using the chain rule,

let, $t=\log x$ and $y=\sin t$

$\because \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}[u \operatorname{sing}$ chain rule]

$\frac{d y}{d x}=\cos t \times \frac{1}{x}$

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

Differentiating again with respect to $x$ :

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\cos (\log \mathrm{x}) \times \frac{1}{\mathrm{x}}\right)$

$\frac{d^{2} y}{d x^{2}}=\cos (\log x) \times \frac{-1}{x^{2}}+\frac{1}{x} \times \frac{1}{x}(-\sin (\log x))$

[ using product rule of differentiation]

$=\frac{-1}{x^{2}} \cos (\log x)-\frac{1}{x^{2}} \sin (\log x)$

$\frac{d^{2} y}{d x^{2}}=\frac{-1}{x^{2}} \cos (\log x)-\frac{1}{x^{2}} \sin (\log x)$