Question

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

Answer 1

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 $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$.