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

$\log (\sin x)$

$\sqrt{B a s i c}$ 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 $f$ 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- $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{uv})=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}$

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

Let's solve now:

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

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

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

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

$\therefore \frac{d y}{d x}=\frac{d}{d x}(\log (\sin x))$

differentiating $\sin (\log x)$ using cthe hain rule,

let, $\mathrm{t}=\sin \mathrm{x}$ and $\mathrm{y}=\log \mathrm{t}$

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

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

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

$\frac{d y}{d x}=\frac{\cos x}{\sin x}=\operatorname{cotx}$

Differentiating again with respect to $x$ :

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(\cot \mathrm{x})$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\operatorname{cosec}^{2} \mathrm{x}\left[\because \frac{\mathrm{d}}{\mathrm{dx}} \cot \mathrm{x}=-\operatorname{cosec}^{2} \mathrm{x}\right]$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\operatorname{cosec}^{2} \mathrm{x}$