Answer
$f'(t)=-te^{t}\csc^{2}t+te^{t}\cot t+e^{t}\cot t$
Work Step by Step
$f(t)=te^{t}\cot t$
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 $te^{t}$:
$...=(te^{t})(-\csc^{2}t)+(\cot t)[(t)(e^{t})'+(e^{t})(t)']=...$
$...=-te^{t}\csc^{2}t+\cot t(te^{t}+e^{t})=-te^{t}\csc^{2}t+te^{t}\cot t+e^{t}\cot t$
