Calculus: Early Transcendentals 8th Edition

$f'(t)=-\dfrac{\csc^{2}t+\cot t}{e^{t}}$
$f(t)=\dfrac{\cot t}{e^{t}}$ 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: $...=\dfrac{e^{t}(-\csc^{2}t-\cot t)}{e^{2t}}=\dfrac{-\csc^{2}t-\cot t}{e^{t}}=-\dfrac{\csc^{2}t+\cot t}{e^{t}}$