Answer
$y'=\dfrac{4e^{2x}}{(1+e^{2x})^{2}}\sin(\dfrac{1-e^{2x}}{1+e^{2x}})$
Work Step by Step
$y=\cos(\dfrac{1-e^{2x}}{1+e^{2x}})$
Differentiate using the chain rule:
$y'=-(\dfrac{1-e^{2x}}{1+e^{2x}})'\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=...$
Use the quotient rule to find $(\dfrac{1-e^{2x}}{1+e^{2x}})'$:
$...=-[\dfrac{(1+e^{2x})(1-e^{2x})'-(1-e^{2x})(1+e^{2x})'}{(1+e^{2x})^{2}}]\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=$
$...=-[\dfrac{-(1+e^{2x})(e^{2x})'-(1-e^{2x})(e^{2x})'}{(1+e^{2x})^{2}}]\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=...$
Use the chain rule to find $(e^{2x})'$:
$...=-[\dfrac{-(1+e^{2x})(2e^{2x})-(1-e^{2x})(2e^{2x})}{(1+e^{2x})^{2}}]\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=...$
Simplify:
$...=-[\dfrac{-2e^{2x}-2e^{4x}-2e^{2x}+2e^{4x}}{(1+e^{2x})^{2}}]\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=...$
$...=\dfrac{4e^{2x}}{(1+e^{2x})^{2}}\sin(\dfrac{1-e^{2x}}{1+e^{2x}})$
