For $$x \gt 1$$, if $$(2x)^{2y} = 4e^{2x-2y}$$, then $$(1 + \log_e 2x)^2 \frac{dy}{dx}$$ is equal to

We are given the functional relation for $$x \gt 1$$

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

First take the natural logarithm (base $$e$$) of both sides. Using the law $$\ln a^{b}=b\ln a,$$ we obtain

$$ \ln\!\bigl((2x)^{2y}\bigr)=\ln\!\bigl(4e^{\,2x-2y}\bigr). $$

Thus

$$ 2y\,\ln(2x)=\ln 4 + 2x-2y. $$

We now differentiate implicitly with respect to $$x$$. Remember that $$y=y(x)$$ is a function of $$x$$, so whenever $$y$$ is differentiated we must multiply by $$\dfrac{dy}{dx}$$ (the chain rule).

Derivative of the left side:

$$ \dfrac{d}{dx}\!\bigl[2y\,\ln(2x)\bigr]=2\frac{dy}{dx}\,\ln(2x)+2y\cdot\dfrac{d}{dx}\!\bigl[\ln(2x)\bigr]. $$

Since $$\dfrac{d}{dx}\bigl[\ln(2x)\bigr]=\dfrac{1}{2x}\cdot 2=\dfrac{1}{x},$$ the left derivative becomes

$$ 2\frac{dy}{dx}\,\ln(2x)+\dfrac{2y}{x}. $$

Derivative of the right side:

$$ \dfrac{d}{dx}\!\bigl[\ln 4 + 2x-2y\bigr]=0+2-2\frac{dy}{dx}. $$

Equating the two derivatives, we have

$$ 2\frac{dy}{dx}\,\ln(2x)+\dfrac{2y}{x}=2-2\frac{dy}{dx}. $$

Gathering the $$\dfrac{dy}{dx}$$ terms on one side:

$$ 2\frac{dy}{dx}\,\ln(2x)+2\frac{dy}{dx}=2-\dfrac{2y}{x}. $$

Factor out $$2\dfrac{dy}{dx}$$:

$$ 2\frac{dy}{dx}\,\bigl(\ln(2x)+1\bigr)=2-\dfrac{2y}{x}. $$

Divide by $$2\bigl(\ln(2x)+1\bigr)$$ to get

$$ \frac{dy}{dx}= \frac{1-\dfrac{y}{x}}{\ln(2x)+1}. $$

Because we ultimately need $$(1+\ln 2x)^2\dfrac{dy}{dx},$$ we next express $$y$$ explicitly in terms of $$x$$. Returning to the original logarithmic equation

$$ 2y\,\ln(2x)+2y=\ln 4+2x, $$

or equivalently

$$ 2y\bigl(\ln(2x)+1\bigr)=\ln 4+2x. $$

Hence

$$ y=\frac{\ln 4+2x}{2\bigl(\ln(2x)+1\bigr)}. $$

Now compute $$1-\dfrac{y}{x}$$:

$$ 1-\frac{y}{x}=1-\frac{\ln 4+2x}{2x\bigl(\ln(2x)+1\bigr)} =\frac{2x\bigl(\ln(2x)+1\bigr)-\bigl(\ln 4+2x\bigr)} {2x\bigl(\ln(2x)+1\bigr)} =\frac{2x\ln(2x)-\ln 4}{2x\bigl(\ln(2x)+1\bigr)}. $$

Therefore

$$ (1+\ln 2x)^2\frac{dy}{dx} =(1+\ln 2x)\Bigl(1-\frac{y}{x}\Bigr) =(1+\ln 2x)\, \frac{2x\ln(2x)-\ln 4}{2x\bigl(\ln(2x)+1\bigr)} =\frac{2x\ln(2x)-\ln 4}{2x}. $$

The factor $$1+\ln 2x$$ in numerator and denominator cancels neatly, leaving

$$ (1+\ln 2x)^2\frac{dy}{dx} =\frac{2x\ln(2x)-\ln 4}{2x} =\ln(2x)-\frac{\ln 2}{x}. $$

Since $$\ln 4=2\ln 2,$$ the final form may be written as

$$ \frac{x\,\ln(2x)-\ln 2}{x}, $$

which coincides exactly with Option B.

Hence, the correct answer is Option B.