Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises - Page 419: 50

$f'(t)=sin2t(e^{sin^{2}t})[sin(2e^{sin^{2}t})]$

$f'(t)=\frac{d}{dt}sin^{2}(e^{sin^{2}t})$ $=2sin(e^{sin^{2}t})\frac{d}{dt}[sin(e^{sin^{2}t})]$ $=2sin(e^{sin^{2}t})cos(e^{sin^{2}t}).(e^{sin^{2}t})\frac{d}{dt}({sin^{2}t)}$ $=sin2(e^{sin^{2}t})(e^{sin^{2}t})(2sintcost)$ Hence,$f'(t)=sin2t(e^{sin^{2}t})[sin(2e^{sin^{2}t})]$