Answer
$f'(t)=-\dfrac{\csc^{2}t+\cot t}{e^{t}}$
Work Step by Step
$f(t)=\dfrac{\cot t}{e^{t}}$
Differentiate using the quotient rule:
$f'(t)=\dfrac{(e^{t})(\cot t)'-(\cot t)(e^{t})'}{(e^{t})^{2}}=...$
$...=\dfrac{(e^{t})(-\csc^{2}t)-(\cot t)(e^{t})}{(e^{t})^{2}}=\dfrac{-e^{t}\csc^{2}t-e^{t}\cot t}{e^{2t}}=...$
Take out common factor $e^{t}$ and simplify:
$...=\dfrac{e^{t}(-\csc^{2}t-\cot t)}{e^{2t}}=\dfrac{-\csc^{2}t-\cot t}{e^{t}}=-\dfrac{\csc^{2}t+\cot t}{e^{t}}$