For x > 1 , if (2x) ^ 2y = 4 e ^ 2x - 2y , then (1 + _ e 2x) ^ 2...

Given, ${\left(2x\right)}^{2y} = 4 . {e}^{2x - 2y}$

Taking natural logarithm on both sides, we get

$2ylo{g}_{e}2x = lo{g}_{e}4 + \left(2x - 2y\right)$

$⇒ 2y = \left(\dfrac{lo{g}_{e}4 + 2x}{1 + lo{g}_{e}2x}\right)$

Differentiating both sides with respect to $x$, we get

$\dfrac{2dy}{dx} = \dfrac{\left(1 + \log_{e}2x\right) . 2 - \left(\log_{e}4 + 2x\right)\dfrac{1}{x}}{{\left(1 + \log_{e}2x\right)}^{2}}$ (Using quotient rule)

$⇒ {\left(1 + \log_{e}2x\right)}^{2}\dfrac{dy}{dx} = \left(\dfrac{x . lo{g}_{e}2x - lo{g}_{e}2}{x}\right)$.