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) $ .