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

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

Taking natural logarithm on both sides, we get

$2y\mathrm{l}\mathrm{o}{\mathrm{g}}_{e}2x=\mathrm{l}\mathrm{o}{\mathrm{g}}_{e}4+\left(2x-2y\right)$

$\Rightarrow 2y=\left(\frac{\mathrm{l}\mathrm{o}{\mathrm{g}}_{e}4+2x}{1+\mathrm{l}\mathrm{o}{\mathrm{g}}_{e}2x}\right)$

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

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

$\Rightarrow {\left(1+{\log }_{e}2x\right)}^2\frac{dy}{dx}=\left(\frac{x.\mathrm{l}\mathrm{o}{\mathrm{g}}_{e}2x-\mathrm{l}\mathrm{o}{\mathrm{g}}_{e}2}{x}\right)$.