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If any function is $f\left( x \right) = g\left( x \right)h\left( x \right)$, then its derivative can be found out by $\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {g\left( x \right)} \right)}}{{dx}}h\left( x \right) + \dfrac{{d\left( {h\left( x \right)} \right)}}{{dx}}g\left( x \right)$, this the basically the product rule.

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

The above equation is not in a differential form, first find the first derivative of the given equation, i.e., differentiate both sides with respect to x, and it can be observed that to differentiate this equation we need to use the product rule.

\[

\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{x^3}\log x} \right)}}{{dx}} \\

\dfrac{{dy}}{{dx}} = 3{x^2}\log x + {x^3}\left( {\dfrac{1}{x}} \right) \\

\dfrac{{dy}}{{dx}} = 3{x^2}\log x + {x^2} \\

\]

So, the first order derivative is \[\dfrac{{dy}}{{dx}} = 3{x^2}\log x + {x^2}\].

Now, find the second order derivative of the above equation, i.e., differentiate both sides with respect to x.

\[

\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{d\left( {3{x^2}\log x} \right)}}{{dx}} + \dfrac{{d\left( {{x^2}} \right)}}{{dx}} \\

\dfrac{{{d^2}y}}{{d{x^2}}} = 6x\log x + 2x \\

\]

So, the second order derivative is \[\dfrac{{{d^2}y}}{{d{x^2}}} = 6x\log x + 2x\].

Now, find the third order derivative of the above equation, i.e., differentiate both sides with respect to x.

\[

\dfrac{{{d^3}y}}{{d{x^3}}} = \dfrac{{d\left( {6x\log x + 2x} \right)}}{{dx}} \\

\dfrac{{{d^3}y}}{{d{x^3}}} = \dfrac{{d\left( {6x\log x} \right)}}{{dx}} + \dfrac{{d\left( {2x} \right)}}{{dx}} \\

\dfrac{{{d^3}y}}{{d{x^3}}} = 6\log x + 2 \\

\]

So, the third order derivative is \[\dfrac{{{d^3}y}}{{d{x^3}}} = 6\log x + 2\].

Now, find the fourth order derivative of the above equation, i.e., differentiate both sides with respect to x.

\[

\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{{d\left( {6\log x + 2} \right)}}{{dx}} \\

\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{{d\left( {6\log x} \right)}}{{dx}} + \dfrac{{d\left( 2 \right)}}{{dx}} \\

\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x} + 0 \\

\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x} \\

\]

So, the fourth order derivative is \[\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x}\].