If the solve the problem

Basic idea:

$\sqrt{S e c o n d}$ 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:

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

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

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

Let's solve now:

As we have to prove: $\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{2}}=2 \cos x \operatorname{cose}^{3} x$

We notice a third order derivative in the expression to be proved so first take the step to find the third order derivative.

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

Let's find $-\frac{d^{3} y}{d x^{3}}$

As $\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right)$

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

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

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

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

$\because \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}$ [using chain rule]

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

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

$\frac{d y}{d x}=\frac{\cos x}{\sin x}=\cot x$

Differentiating again with respect to $x$ :

$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(\cot x)$

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

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

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

Differentiating again with respect to $x$ :

$\frac{d}{d x}\left(\frac{d^{2} y}{d x^{2}}\right)=\frac{d}{d x}\left(-\operatorname{cosec}^{2} x\right)$

using the chain rule and $\frac{d}{d x} \operatorname{cosec} x=-\operatorname{cosec} x \cot x$

$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}=-2 \operatorname{cosec} \mathrm{x}(-\operatorname{cosec} \mathrm{x} \cot \mathrm{x})$

$=2 \operatorname{cosec}^{2} x \cot x=2 \operatorname{cosec}^{2} x \frac{\cos x}{\sin x}[\because \cot x=\cos x / \sin x]$

$\therefore \frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}=2 \operatorname{cosec}^{3} \mathrm{x} \cos \mathrm{x} \ldots \ldots . \mathrm{proved}$