If the solve the problem

Basic idea:

$\sqrt{S e c o n d}$ order derivative is nothing but derivative of derivative i.e. $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$

$\sqrt{T h e}$ 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:

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

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 $: \frac{\mathrm{d}^{4} \mathrm{y}}{\mathrm{dx}^{4}}=\frac{6}{\mathrm{x}}$

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

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

Let's find $-\frac{d^{4} y}{d x^{4}}$

As $\frac{\mathrm{d}^{4} \mathrm{y}}{\mathrm{dx}^{4}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{3}}\right)=\frac{\mathrm{d}}{\mathrm{dx}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right)\right)$

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

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

differentiating using product rule:

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

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

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

$\frac{d y}{d x}=x^{2}(1+3 \log x)$

Again differentiating using product rule:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{x}^{2} \frac{\mathrm{d}}{\mathrm{dx}}(1+3 \log \mathrm{x})+(1+3 \log \mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{2}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{x}^{2} \times \frac{3}{\mathrm{x}}+(1+3 \log \mathrm{x}) \times 2 \mathrm{x}$

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

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{x}(5+6 \log \mathrm{x})$

Again differentiating using product rule:

$\frac{d^{3} y}{d x^{3}}=x \frac{d}{d x}(5+6 \log x)+(5+6 \log x) \frac{d}{d x} x$

$\frac{d^{3} y}{d x^{3}}=x \times \frac{6}{x}+(5+6 \log x)$

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

$\frac{d^{3} y}{d x^{3}}=11+6 \log x$

Again differentiating w.r.t $x$ :

$\frac{\mathrm{d}^{4} \mathrm{y}}{\mathrm{dx}^{4}}=\frac{6}{\mathrm{x}} \ldots \ldots$ proved