Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 453: 51

$y'=sin(\frac{1-e^{2x}}{1+e^{2x}})[\frac{4e^{2x}}{1+e^{2x})^{2}}]$

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