The differentiation \[\dfrac{d}{{dx}}\left[ {\log \left\{ {{e^x}{{\left( {\dfrac{{\left( {x - 2} \right)}}{{\left( {x + 2} \right)}}} \right)}^{\dfrac{3}{4}}}} \right\}} \right] \] equal to
(A) \[1\]
(B) \[\dfrac{{\left( {{x^2} + 1} \right)}}{{\left( {{x^2} - 4} \right)}}\]
(C) \[\dfrac{{\left( {{x^2} - 1} \right)}}{{\left( {{x^2} - 4} \right)}}\]
(D) \[{e^x}\dfrac{{\left( {{x^2} - 1} \right)}}{{\left( {{x^2} - 4} \right)}}\]

Hint: This question needs knowledge of derivatives and formulas related to it, like, derivative of \[\log (x)\] is \[\dfrac{1}{x}\] , derivative of x is $1$. We should also remember the basic properties of logarithmic function such as \[\log (e) = 1\] , \[\log \left( {{a^n}} \right) = n\log \left( a \right)\] , \[\log \left( {ab} \right) = \log \left( a \right) + \log \left( b \right)\] and \[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\] . Keep in mind the algebraic property \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\] to get to the final answer.

Complete step-by-step answer:
First we let the function as y so that we can write it easily,
Therefore, we have,
 \[y = \left[ {\log \left\{ {{e^x}{{\left( {\dfrac{{\left( {x - 2} \right)}}{{\left( {x + 2} \right)}}} \right)}^{\dfrac{3}{4}}}} \right\}} \right] \]
Now we apply the logarithmic property \[\log \left( {ab} \right) = \log \left( a \right) + \log \left( b \right)\]
 \[ \Rightarrow y = \log \left[ {{e^x}} \right] + \log \left[ {{{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)}^{\dfrac{3}{4}}}} \right] \]
Now, we apply the property that logarithm \[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\] , we get,
 \[ \Rightarrow y = \log \left[ {{e^x}} \right] + \left[ {\log \left[ {{{\left( {x - 2} \right)}^{\dfrac{3}{4}}}} \right] - \log \left[ {{{\left( {x + 2} \right)}^{\dfrac{3}{4}}}} \right] } \right] \]
Now, applying the power rule of logarithm, that is \[\log \left( {{a^n}} \right) = n\log \left( a \right)\] , we get,
 \[ \Rightarrow y = \log \left[ {{e^x}} \right] + \left[ {\dfrac{3}{4}\log \left( {x - 2} \right) - \dfrac{3}{4}\log \left( {x + 2} \right)} \right] \]
We know that, natural logarithm of e is $1$ so we get,
 \[ \Rightarrow y = x\log e + \left[ {\dfrac{3}{4}\log \left( {x - 2} \right) - \dfrac{3}{4}\log \left( {x + 2} \right)} \right] \]
 \[ \Rightarrow y = x + \dfrac{3}{4}\log \left( {x - 2} \right) - \dfrac{3}{4}\log \left( {x + 2} \right)\]
Now, differentiating both sides with respect to x,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {x + \dfrac{3}{4}\log \left( {x - 2} \right) - \dfrac{3}{4}\log \left( {x + 2} \right)} \right] \]
Applying the addition property of derivative, that is, \[\dfrac{d}{{dx}}[u + v] = \dfrac{{du}}{{dx}} + \dfrac{{dv}}{{dx}}\]
So, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ x \right] + \dfrac{d}{{dx}}\left[ {\dfrac{3}{4}\log \left( {x - 2} \right)} \right] - \dfrac{d}{{dx}}\left[ {\dfrac{3}{4}\log \left( {x + 2} \right)} \right] \]
Now, we know that the derivative of x is $1$ and we can take the constant out of the differentiation. Now, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = 1 + \dfrac{3}{4}\left( {\dfrac{{1\left( 1 \right)}}{{x - 2}}} \right) - \dfrac{3}{4}\left( {\dfrac{{1\left( 1 \right)}}{{x + 2}}} \right)\]
On taking LCM in the denominator we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4\left( {x - 2} \right)\left( {x + 2} \right) + 3\left( {x + 2} \right) - 3\left( {x - 2} \right)}}{{4\left( {x - 2} \right)\left( {x + 2} \right)}}\]
Applying the algebraic property \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\] , we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4\left( {{x^2} - 4} \right) + 3x + 6 - 3x + 6}}{{4\left( {{x^2} - 4} \right)}}\]
Now, on solving we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4\left( {{x^2} - 4} \right) + 12}}{{4\left( {{x^2} - 4} \right)}}\]
Taking $4$ as common in denominator,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4\left( {{x^2} - 4 + 3} \right)}}{{4\left( {{x^2} - 4} \right)}}\]
Cancelling common factors in numerator and denominator, we get,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{x^2} - 1} \right)}}{{\left( {{x^2} - 4} \right)}}\]
Hence, option (C) is the correct answer.

Note: This question requires knowledge of formulas of derivatives like derivative of \[\log (x)\] is \[\dfrac{1}{x}\] , derivative of x is 1. We should also remember the basic logarithmic properties, that is, \[\log (e) = 1\] , \[\log \left( {{a^n}} \right) = n\log \left( a \right)\] , \[\log \left( {ab} \right) = \log \left( a \right) + \log \left( b \right)\] and \[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\] by heart. Algebraic property \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\] should also be kept in mind. Take care while doing the calculations.