Question

For ${x^2} - 4 \ne 0$ , the value of \dfrac{d}{{dx}}\left[ {\log \left\\{ {{e^x}{{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)}^{\dfrac{3}{4}}}} \right\\}} \right] at $x = 3$ is:
A. $1$
B. $\dfrac{8}{5}$
C. $2$
D. $\dfrac{{8{e^3}}}{5}$

Answer 1

Hint : First simplify the given function with the help of the properties of logarithm and then differentiate it with respect to the independent variable $x$ . Substitute the value of x=3 in the final obtain equation after differentiating to get the required value.

Complete step-by-step answer :
Simplify the function given in the question with the help of properties of logarithm:
\log \left\\{ {{e^x}{{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)}^{\dfrac{3}{4}}}} \right\\} = \log \left( {{e^x}} \right) + \log \left\\{ {{{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)}^{\dfrac{3}{4}}}} \right\\} \\\ = x\log e + \dfrac{3}{4}\log \left( {\dfrac{{x - 2}}{{x + 2}}} \right) \\\ = x\left( 1 \right) + \dfrac{3}{4}\left( {\log \left( {x - 2} \right) - \log \left( {x + 2} \right)} \right) \\\ = x + \dfrac{3}{4}\log \left( {x - 2} \right) - \dfrac{3}{4}\log \left( {x + 2} \right) \;
Now differentiate the expression with respect to the independent variable $x$ .
\dfrac{d}{{dx}}\left[ {\log \left\\{ {{e^x}{{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)}^{\dfrac{3}{4}}}} \right\\}} \right] = \dfrac{d}{{dx}}\left( {x + \dfrac{3}{4}\log \left( {x - 2} \right) + \dfrac{3}{4}\log \left( {x + 2} \right)} \right) \\\ = \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) \\\ = 1 + \dfrac{3}{4} \times \dfrac{1}{{x - 2}} \times \dfrac{d}{{dx}}\left( {x - 2} \right) - \dfrac{3}{4} \times \dfrac{1}{{x + 2}} \times \dfrac{d}{{dx}}\left( {x + 2} \right) \\\ = 1 + \dfrac{3}{{4\left( {x - 2} \right)}} \times 1 - \dfrac{3}{{4\left( {x + 2} \right)}} \times 1 \\\ = 1 + \dfrac{3}{{4\left( {x - 2} \right)}} - \dfrac{3}{{4\left( {x + 2} \right)}} \;
Further simplify,
\dfrac{d}{{dx}}\left[ {\log \left\\{ {{e^x}{{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)}^{\dfrac{3}{4}}}} \right\\}} \right] = 1 + \dfrac{3}{{4\left( {x - 2} \right)}} - \dfrac{3}{{4\left( {x + 2} \right)}} \\\ = \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)}} \\\ = \dfrac{{4\left( {{x^2} - 4} \right) + 3x + 6 - 3x + 6}}{{4\left( {{x^2} - 4} \right)}} \\\ = \dfrac{{4{x^2} + 12 - 16}}{{4\left( {{x^2} - 4} \right)}} \\\ = \dfrac{{{x^2} - 1}}{{{x^2} - 4}} \;
Now substitute $3$ for $x$ in the expression to get the value of \dfrac{d}{{dx}}\left[ {\log \left\\{ {{e^x}{{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)}^{\dfrac{3}{4}}}} \right\\}} \right] at $x = 3$ .
\dfrac{d}{{dx}}{\left[ {\log \left\\{ {{e^x}{{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)}^{\dfrac{3}{4}}}} \right\\}} \right]_{x = 3}} = \dfrac{{{{\left( 3 \right)}^2} - 1}}{{{{\left( 3 \right)}^2} - 4}} \\\ = \dfrac{{9 - 1}}{{9 - 4}} \\\ = \dfrac{8}{5} \;
So, the correct answer is “Option B”.

Note : Simplify the function first and then solve otherwise it will be a complicated problem to solve. The logarithm properties used in this question are as follows:
$\log \left( {ab} \right) = \log a + \log b \\\ \log \left( {\dfrac{a}{b}} \right) = \log a - \log b \\\ \log {a^b} = b\log a \;$
The differentiation of a function $g\left( x \right)$ which is equal to $\log \left( {f\left( x \right)} \right)$ is given by $\dfrac{d}{{dx}}\left( {g\left( x \right)} \right)$ which is equal to $\dfrac{1}{{f\left( x \right)}} \times \dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$ .