How do you differentiate $f(x) = {x^3}\ln x$ using the product rule?

Hint: Here we have to find out the derivative of the given term by using to solve the product rule. Then, we will rearrange the given expression and use the product rule to find the derivative of the equation. Finally we get the required answer.

Formula used: $\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$

Complete step-by-step solution:
We have the given equation as:
$f(x) = {x^3}\ln x$
And we have to find the derivative of this equation therefore, it can be written as:
$ \Rightarrow \dfrac{d}{{dx}}{x^3}\ln x$
Now since the expression is in the form of multiplication of two terms which are ${x^3}$ and $\ln x$, we will use the product formula which is:$\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
Now on considering $u = {x^3}$ and $v = \ln x$ and applying the formula, we get:
$ \Rightarrow \dfrac{d}{{dx}}({x^3}\ln x) = {x^3}\dfrac{d}{{dx}}\ln x + \ln x\dfrac{d}{{dx}}{x^3}$
Now we know the formula that $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$ , on applying these formulas in the term, we get:
$ \Rightarrow \dfrac{d}{{dx}}({x^3}\ln x) = {x^3} \times \dfrac{1}{x} + \ln x \times 3{x^{3 - 1}}$
On simplifying the equation by cancelling $x$, we get:
$ \Rightarrow \dfrac{d}{{dx}}({x^3}\ln x) = {x^2} + \ln x \times 3{x^2}$
On rearranging the terms, we get:
$ \Rightarrow \dfrac{d}{{dx}}({x^3}\ln x) = {x^2} + 3{x^2}\ln x$

The answer for the given question is $\dfrac{d}{{dx}}({x^3}\ln x) = {x^2} + 3{x^2}\ln x$.

Note: In this question we have used the product formula; there also exists the division formula which is:
$\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}$ , which should be remembered to do the division sums
To do these types of questions all the basic derivative formulas should be remembered.
It is to be remembered that integration and derivative are the inverse of each other, if the derivative of $a$ is $b$, then the integration of $b$ will be $a$.
If there is a term which has more than to functions of derivatives which means that it is the form of $f(g(x))$, then the chain rule has to be used which is $F'(x) = f'(g(x))g'(x)$, this should be done till the point the equation is in the simple format.
The term $\ln x$ in the equation represents the natural log which has base $e$.