Find \[y'\] and $y''$? $y = {x^2}\ln \left( {2x} \right)$

Hint: First of all we have to convert the function in the form of a product of functions. Then we will apply the formula of differentiation of the product of functions, which is ${\left( {u.v} \right)^\prime } = u'.v + u.v'$. This will give us the first derivative of the function, that is $y'$. Then we will again differentiate the first derivative using the required formulas and methods. Then we will get the second derivative of the given function. So, let us see how to solve the problem.

Complete step by step answer:
The given function is, $y = {x^2}\ln \left( {2x} \right)$. Therefore, we will differentiate the function using the product rule of derivative.
\[y' = \dfrac{d}{{dx}}\left( {{x^2}\ln \left( {2x} \right)} \right)\]
By using the product rule, that gives, ${\left( {u.v} \right)^\prime } = u'.v + u.v'$ and also the chain rule simultaneously, we get,
\[ \Rightarrow y' = \dfrac{d}{{dx}}\left( {{x^2}} \right).\ln \left( {2x} \right) + {x^2}.\dfrac{d}{{dx}}\left( {\ln \left( {2x} \right)} \right)\]
\[ \Rightarrow y' = 2x.\ln \left( {2x} \right) + {x^2}.\dfrac{1}{{2x}}.2\]

Now, by simplifying, we get,
\[ \Rightarrow y' = 2x\ln \left( {2x} \right) + x\]
Therefore, for the second derivative of the function, we will again differentiate the function with respect to $x$.Therefore,
$y'' = \dfrac{d}{{dx}}y'$
$ \Rightarrow y'' = \dfrac{d}{{dx}}\left( {2x\ln \left( {2x} \right) + x} \right)$
Now, separately differentiating each term separated by the sum operation, we get,
$ \Rightarrow y'' = \dfrac{d}{{dx}}\left( {2x\ln \left( {2x} \right)} \right) + \dfrac{d}{{dx}}x$

Now, again using the product rule of differentiation and the chain rule simultaneously, we get,
$ \Rightarrow y'' = \left[ {2.\ln \left( {2x} \right) + 2x.\dfrac{1}{{2x}}.2} \right] + 1$
Now, simplifying the terms, we get,
$ \Rightarrow y'' = 2\ln \left( {2x} \right) + 2 + 1$
$ \Rightarrow y'' = 2\ln \left( {2x} \right) + 3$
Therefore, the first and second derivatives of the given function are,
\[y' = 2x\ln \left( {2x} \right) + x\]
$\therefore y'' = 2\ln \left( {2x} \right) + 3$

Therefore, the value of $y''$ is $2\ln \left( {2x} \right) + 3$.

Note: The derivatives of basic trigonometric and algebraic functions must be learned by heart in order to find derivatives of complex composite functions using product rule and chain rule of differentiation. The product rule of differentiation involves differentiating a product of two or more functions and the chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.