Find : \[\dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = \]
A. \[{e^x}\left( {\log \sin 2x + 2\cot 2x} \right)\]
B. \[{e^x}\left( {\log \cos 2x + 2\cot 2x} \right)\]
C. \[{e^x}\left( {\log \cos 2x + \cot 2x} \right)\]
D.None of these.

Hint: Here, the given question. We have to find the derivative or differentiated term of the function. For this, first we need to differentiate the given function with respect to \[x\] by using a standard differentiation formula then by using product and chain rule for differentiation then on further simplification we get the required differentiation value.

Complete step-by-step answer:
Differentiation can be defined as a derivative of a function with respect to an independent variable
Otherwise
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Let \[y = f\left( x \right)\] be a function of. Then the rate of change of “y” per unit change in “x” is given by \[\dfrac{{dy}}{{dx}}\] .
The Chain Rule is a formula for computing the derivative of the composition of two or more
functions.
The chain rule expressed as \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]
The product rule is used to differentiate many functions where one function is multiplied by another. The formal definition of the rule is: \[\dfrac{d}{{dx}}\left( {uv} \right) = u \cdot \dfrac{{dv}}{{dx}} + v \cdot \dfrac{{du}}{{dx}}\]
Consider the given function
  \[ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right)\] ---------- (1)
Now we have to differentiate this function with respect to \[x\] by applying a product rule of differentiation, then
Here, \[u = {e^x}\] and \[v = \log \sin 2x\]
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = {e^x} \cdot \dfrac{d}{{dx}}\left( {\log \sin 2x} \right) + \log \sin 2x \cdot \dfrac{d}{{dx}}\left( {{e^x}} \right)\] -----(2)
Using the standard differentiated formula \[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\] and \[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\] , then equation (2) becomes
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = {e^x} \cdot \dfrac{1}{{\sin 2x}}\dfrac{d}{{dx}}\left( {\sin 2x} \right) + \log \sin 2x \cdot {e^x}\]
As we know another standard differentiated formula \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\] , then
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = {e^x} \cdot \dfrac{1}{{\sin 2x}}\left( {\cos 2x} \right)\dfrac{d}{{dx}}\left( {2x} \right) + \log \sin 2x \cdot {e^x}\]
Again, apply a differentiated formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] , then we have
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = {e^x} \cdot \dfrac{{\cos 2x}}{{\sin 2x}}\left( 2 \right) + \log \sin 2x \cdot {e^x}\]
By the definition of trigonometric ratio cot is a ratio of cos and sin i.e., \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] , then
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = 2{e^x} \cdot \cot 2x + \log \sin 2x \cdot {e^x}\]
Take \[{e^x}\] as common
 \[\therefore \dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = {e^x}\left( {2\cot 2x + \log \sin 2x} \right)\]
Hence, the required the differentiated value \[\dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = {e^x}\left( {2\cot 2x + \log \sin 2x} \right)\] .
Therefore, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note: When differentiating the function or term the student must recognize the dependent and independent variable then differentiate the dependent variable with respect to the independent variable. Should know the product and quotient rule of differentiation and remember the standard differentiation formulas of logarithm, trigonometric and exponential functions.