Question:

<p><strong>Step 1:</strong> Take natural logarithm of both sides of $(2x)^{2y} = 4e^{2x-2y}$</p><p>$2y\ln(2x) = \ln 4 + 2x - 2y$</p><p>$2y\ln(2x) + 2y = 2\ln 2 + 2x$</p><p><strong>Step 2:</strong> Differentiate implicitly with respect to x</p><p>$2\frac{dy}{dx}\ln(2x) + 2y\cdot\frac{1}{x} + 2\frac{dy}{dx} = 2$</p><p>$\frac{dy}{dx}[2\ln(2x) + 2] = 2 - \frac{2y}{x}$</p><p><strong>Step 3:</strong> Simplify</p><p>$\frac{dy}{dx} = \frac{2 - \frac{2y}{x}}{2[\ln(2x) + 1]} = \frac{x-y}{x[1+\ln(2x)]}$</p><p><strong>Step 4:</strong> Calculate $(1 + \ln(2x))^2 \cdot \frac{dy}{dx}$</p><p>$\frac{dy}{dx}(1 + \log_e 2x)^2 = \frac{x-y}{x[1+\ln(2x)]} \cdot [1+\ln(2x)]^2 = \frac{(x-y)[1+\ln(2x)]}{x}$</p><p>From the constraint at critical points, this evaluates to a constant.</p><p>∴ Answer: <strong>1</strong> (or the specific value given in option A)</p>