Question

For $x > 1$, if $(2x)^{2y} = 4e^{2x - 2y}$, then $\left( 1 +\log_{e} 2x\right)^{2} \frac{dy}{dx} $ is equal to :

• A. $\log_e 2x$
• B. $\frac{x \log_{e} 2x +\log_{e} 2}{x} $
• C. $x \log_e 2x$
• D. $\frac{x \log_{e} 2x - \log_{e} 2}{x} $

Answer 1

Answer: (D)

$(2x)^{2y} = 4e^{2x - 2y}$
$2y \ell n2x = \ell n4 + 2x - 2y$
$y = \frac{x + \ell n 2}{1 + \ell n 2 x}$
$y' = \frac{(1 + \ell n 2 x) - (x + \ell n 2 ) \frac{1}{x}}{( 1+ \ell n 2x )^2}$
$y ' = 1 + \ell n 2 x)^2 = [\frac{x \ell n 2x - \ell n 2}{x} ] $