For \(x>1\), if \((2 x)^{2 y}=4 e^{2 x-2 y}\), then \((1+\log 2 x)^2 \frac{d y}{d x}\) is equal to

\(\begin{aligned} & (2 x)^{2 y}=4 e^{2 x-2 y} \Rightarrow 2 y \log (2 x)=\log 4+2 x-2 y \\ & \Rightarrow 2 y \log (2 x)=2 \log 2+2 x-2 y\end{aligned}\)

Different w.r.t \(x\)
\(\begin{aligned} & \frac{d y}{d x} \cdot \log (2 x)+y \cdot \frac{1}{2 x} \cdot 2=0+1-\frac{d y}{d x} \\ & \Rightarrow \frac{d y}{d x}(1+\log 2 x)=1-\frac{y}{x}=1-\frac{x+\log 2}{x(1+\log 2 x)} \\ & {\left[\because \text { from }(1) y=\frac{x+\log 2}{(1+\log 2 x)}\right]} \\ & \Rightarrow \frac{d y}{d x}(1+\log 2 x)^2=\frac{x(1+\log 2 x)-x-\log 2}{x} \\ & \Rightarrow(1+\log 2 x)^2=\frac{d y}{d x}=\frac{x \log 2 x-\log 2}{x}\end{aligned}\)