Answer
$F'(t)=(e^{t\sin2t})(2t\cos2t+\sin2t)$
Work Step by Step
$F(t)=e^{t\sin2t}$
Differentiate using the chain rule:
$F'(t)=(e^{t\sin2t})(t\sin2t)'=...$
Apply the product rule to find $(t\sin2t)'$:
$...=(e^{t\sin2t})[(t)(\sin2t)'+(\sin2t)(t)']=...$
Apply the chain rule one more time to find $(\sin2t)'$:
$...=(e^{t\sin2t})[(t)(\cos2t)(2t)'+(\sin2t)(1)]=...$
Simplify:
$...=(e^{t\sin2t})(2t\cos2t+\sin2t)$
