How do you differentiate $$x{\left( {\ln x} \right)^2}$$?

Hint: In solving the question, differentiate the given equation by using product rule i.e, $$\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right]$$ is $$f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]$$ and the result expression must be again differentiated by using the chain rule, which states that $$\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$$,then we will get the required result.

Complete step-by-step answer:
Given expression is $$x{\left( {\ln x} \right)^2}$$,
Differentiating using the product rule which states that, $$\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right]$$ is $$f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]$$ where $$f\left( x \right) = x$$ and $$g\left( x \right) = {\left( {\ln x} \right)^2}$$,
The expression becomes,
$$ \Rightarrow x\dfrac{d}{{dx}}\left[ {{{\left( {\ln x} \right)}^2}} \right] + {\left( {\ln x} \right)^2}\dfrac{d}{{dx}}x$$,
Differentiate using the chain rule, which states that $$\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$$ where nmn$$f\left( x \right) = {x^2}$$ and $$g\left( x \right) = \ln x$$,
To apply chain rule set $$u$$ as $$\ln x$$,
We know that $$\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$$, now the expression becomes,
nnn$$ \Rightarrow x\left( {\dfrac{d}{{du}}\left[ {{u^2}} \right]\dfrac{d}{{dx}}\left( {\ln x} \right)} \right) + {\left( {\ln x} \right)^2}\dfrac{d}{{dx}}x$$,
we know that $$\dfrac{d}{{dx}}x = 1$$, now the expression becomes, as
$$ \Rightarrow x\left[ {2u \cdot \dfrac{1}{x}} \right] + {\left( {\ln x} \right)^2} \cdot 1$$,
Now substitute the value of u we get,
$$ \Rightarrow x\left[ {2\left( {\ln x} \right) \cdot \dfrac{1}{x}} \right] + {\left( {\ln x} \right)^2} \cdot 1$$,
Now simplifying by eliminating the like terms we get,
$$ \Rightarrow 2\ln x + {\left( {\ln x} \right)^2}$$
So the derivative of the given expression $$x{\left( {\ln x} \right)^2}$$ is $$2\ln x + {\left( {\ln x} \right)^2}$$.
Final Answer:

$$\therefore $$ The differentiation value of $$x{\left( {\ln x} \right)^2}$$ is $$2\ln x + {\left( {\ln x} \right)^2}$$

Note:
Differentiation is the method of evaluating a function’s derivative at any time. Some of the fundamental rules for differentiation are given below, and using these rules we can solve differentiation questions easily.
Sum or difference rule: When the function is the sum or difference of two functions, the derivative is the sum or difference of derivative of each function, i.e.,
If $$f\left( x \right) = u\left( x \right) \pm v\left( x \right)$$,
Then $$f'\left( x \right) = u'\left( x \right) \pm v'\left( x \right)$$.
Product rule: When the function is the product of two functions, the derivative is the sum or difference of derivative of each function, i.e.,
If $$f\left( x \right) = u\left( x \right) \times v\left( x \right)$$,
Then $$f'\left( x \right) = u'\left( x \right)v\left( x \right) + u\left( x \right)v'\left( x \right)$$
Quotient Rule: If the function is in the form of two functions $$\dfrac{{u\left( x \right)}}{{v\left( x \right)}}$$, the derivative of the function can be expressed as,
$$f'\left( x \right) = \dfrac{{u'\left( x \right)v\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{{\left( {v\left( x \right)} \right)}^2}}}$$.
Chain Rule:
If $$y = f\left( x \right) = g\left( u \right)$$,
And if $$u = h\left( x \right)$$,
Then $$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$$.