## Calculus: Early Transcendentals 8th Edition

$f'(t)=-te^{t}\csc^{2}t+te^{t}\cot t+e^{t}\cot t$
$f(t)=te^{t}\cot t$ This is a product of three functions. To differentiate it, we will apply the product rule taking $te^{t}$ as one function. Then we will need to use the product rule a second time to finish the problem. $f'(t)=(te^{t})(\cot t)'+(\cot t)(te^{t})'=...$ Use the product rule to differentiate $te^{t}$: $...=(te^{t})(-\csc^{2}t)+(\cot t)[(t)(e^{t})'+(e^{t})(t)']=...$ $...=-te^{t}\csc^{2}t+\cot t(te^{t}+e^{t})=-te^{t}\csc^{2}t+te^{t}\cot t+e^{t}\cot t$