Answer
$y'=2e^{\sin2x}\cos2x+2e^{2x}\cos(e^{2x})$
Work Step by Step
$y=e^{\sin2x}+\sin(e^{2x})$
Differentiate each term:
$y'=(e^{\sin2x})'+[\sin(e^{2x})]'=...$
Apply the chain rule to find $(e^{\sin2x})'$ and $[\sin(e^{2x})]'$:
$...=(e^{\sin2x})(\sin2x)'+[\cos(e^{2x})](e^{2x})'=...$
Use the chain rule one more time to find $(\sin2x)'$ and $(e^{2x})'$:
$...=(e^{\sin2x})(\cos2x)(2x)'+[\cos(e^{2x})](e^{2x})(2x)'=...$
$...=2e^{\sin2x}\cos2x+2e^{2x}\cos(e^{2x})$
