Question

If $y = {x^3}\log x$, prove that $\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x}$.

Answer 1

Hint- As, it is given in the equation is not in differential form, so to find the fourth derivative we need to differentiate it four times, i.e., we have to find the first, second, third and then the fourth derivative, by doing this we will get the desired equation. Here in this equation we have to observe one thing that the given equation we can follow the product rule to differentiate it

Formula used:
 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.


Complete step by step solution:
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 \[\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x}\].
Hence, proved $\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x}$.


Note- The given equation is $y = {x^3}\log x$, as it can be observed that the order and degree of the differential equation is null as it is a simple equation, and it is asked to find the fourth order and one degree differential equation. So, it is obvious that we need to differentiate it four times. Take care to see that the equation is differentiated exactly as many as the number of times (degree) as asked in the question