## Calculus: Early Transcendentals 8th Edition

$V'(t)=-\dfrac{t^{2}+4t+4}{t^{2}e^{t}}$
$V(t)=\dfrac{4+t}{te^{t}}$ Differentiate using the quotient rule: $V'(t)=\dfrac{(te^{t})(4+t)'-(4+t)(te^{t})'}{(te^{t})^{2}}=...$ Evaluate $(te^{t})'$ using the product rule: $...=\dfrac{(te^{t})(1)-(4+t)[(t)(e^{t})'+(e^{t})(t)']}{t^{2}e^{2t}}=...$ $...=\dfrac{te^{t}-(4+t)(te^{t}+e^{t})}{t^{2}e^{2t}}=...$ Simplify: $...=\dfrac{te^{t}-4te^{t}-4e^{t}-t^{2}e^{t}-te^{t}}{t^{2}e^{2t}}=\dfrac{-t^{2}e^{t}-4te^{t}-4e^{t}}{t^{2}e^{2t}}=...$ Take out common factor $e^{t}$ and continue simplifying: $...=\dfrac{e^{t}(-t^{2}-4t-4)}{t^{2}e^{2t}}=\dfrac{-t^{2}-4t-4}{t^{2}e^{t}}=-\dfrac{t^{2}+4t+4}{t^{2}e^{t}}$