Answer
$y'=sin(\frac{1-e^{2x}}{1+e^{2x}})[\frac{4e^{2x}}{1+e^{2x})^{2}}]$
Work Step by Step
$y'=\frac{d}{dx}[cos(\frac{1-e^{2x}}{1+e^{2x}})]$
$=-sin(\frac{1-e^{2x}}{1+e^{2x}})\frac{d}{dx}(\frac{1-e^{2x}}{1+e^{2x}})$
$=-sin(\frac{1-e^{2x}}{1+e^{2x}})[\frac{-4e^{2x}}{1+e^{2x})^{2}}]$
Hence, $y'=sin(\frac{1-e^{2x}}{1+e^{2x}})[\frac{4e^{2x}}{1+e^{2x})^{2}}]$