Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 26

Differentiate $\displaystyle y = e^{-t} (t^2 - 2t + 2)$


$
\begin{equation}
\begin{aligned}

y =& \frac{t^2 - 2t + 2}{e^t}
\\
\\
y' =& \frac{d}{dt} \left( \frac{t^2 - 2t + 2}{e^t} \right)
\\
\\
y' =& \frac{\displaystyle e^t \frac{d}{dt} (t^2 - 2t + 2) - (t^2 - 2t + 2) \frac{d}{dt} (e^t)}{(e^t)^2}
\\
\\
y' =& \frac{(e^t) (2t -2) - (t^2 - 2t + 2)(e^t) }{e^{2t}}
\\
\\
y' =& \frac{e^t (2t - 2 - t^2 + 2t - 2)}{e^{2t}}
\\
\\
y' =& \frac{-t^2 + 4t - 4}{e^t}
\\
\\
& \text{ or }
\\
\\
y' =& e^{-t} (-t^2 + 4t - 4)


\end{aligned}
\end{equation}
$