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

\((2 x)^{2 y}=4 \mathrm{e}^{2 x-2 y}\)
Taking log on both sides, we get
\(2 y \log 2 x=\log \left(4 \mathrm{e}^{2 x-2 y}\right) \)
\( \Rightarrow 2 y \log 2 x=\log 4+\log \mathrm{e}^{2 x-2 y} \)
\( \Rightarrow 2 y \log 2 x=\log 4+2 x-2 y \)
\( \Rightarrow 2 y \log 2 x+2 y=\log 4+2 x \)
\( \Rightarrow 2(y \log 2 x+y)=2 \log 2+2 x \)
\( \Rightarrow y \log 2 x+y=\log 2+x \)
\( \Rightarrow y(1+\log 2 x)=x+\log 2 \)
\( \Rightarrow y=\frac{x+\log 2}{1+\log 2 x} \)
Differentiating w.r.t. \(x\), we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(1+\log 2 x)(1+0)-(x+\log 2)\left(\frac{1}{2 x}\right) \cdot 2}{(1+\log 2 x)^2} \)
\( \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(1+\log 2 x)-\frac{1}{x}(x+\log 2)}{(1+\log 2 x)^2} \)
\( \Rightarrow(1+\log 2 x)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}=1+\log 2 x-1-\frac{\log 2}{x} \)
\( \Rightarrow(1+\log 2 x)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}=\log 2 x-\frac{\log 2}{x} \)
\( =\frac{x \log 2 x-\log 2}{x}\)