Question:
For $x>1$, if $(2 x)^{2 y}=4 e^{2 x-2 y}$, then
$\left(1+\log _{e} 2 x\right)^{2} \frac{d y}{d x}$ is equal to :
Correct Option: , 4
Solution:
$(2 x)^{2 y}=4 e^{2 x-2 y}$
$2 \mathrm{y} \ell \mathrm{n} 2 \mathrm{x}=\ell \mathrm{n} 4+2 \mathrm{x}-2 \mathrm{y}$
$y=\frac{x+\ell n 2}{1+\ell n 2 x}$
$y^{\prime}=\frac{(1+\ell \operatorname{n} 2 x)-(x+\ell n 2) \frac{1}{x}}{(1+\ell n 2 x)^{2}}$
$y^{\prime}(1+\ell n 2 x)^{2}=\left[\frac{x \ell n 2 x-\ell n 2}{x}\right]$