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

\(
(2 x)^{2 y}=4 \mathrm{e}^{2 x-2 y}
\)
Taking ' \(\log _{\mathrm{e}}\) ' on both sides, we get
\(
2 y \log _{\mathrm{e}}(2 x)=\log _{\mathrm{e}} 4+(2 x-2 y) \log _{\mathrm{e}} \mathrm{e}
\)
\(
\therefore y \log _{\mathrm{e}}(2 x)=\log _{\mathrm{e}} 2+x-y
\)
Differentiating w.r.t. \(x\), we get
\({\left[\log _{\mathrm{e}} 2 x\right] \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 y}{2 x}=0+1-\frac{\mathrm{d} y}{\mathrm{~d} x}} \)
\( {\left[1+\log _{\mathrm{e}} 2 x\right] \frac{\mathrm{d} y}{\mathrm{~d} x}=1-\frac{y}{x}} \)
\( \text { Now, (i) } \Rightarrow y=\frac{\log _{\mathrm{e}} 2+x}{1+\log _{\mathrm{e}} 2 x} \)
\( \therefore \ldots \text { (ii) } \left(\text { ii) } \Rightarrow\left(1+\log _{\mathrm{e}} 2 x\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x \log _{\mathrm{e}} 2 x-\log _{\mathrm{e}} 2}{x\left(1+\log _e 2 x\right)}\right. \)
\( \therefore\left(1+\log _{\mathrm{e}} 2 x\right)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x \log _{\mathrm{e}} 2 x-\log _{\mathrm{e}} 2}{x}\)