Answer
$V'(t)=-\dfrac{t^{2}+4t+4}{t^{2}e^{t}}$
Work Step by Step
$V(t)=\dfrac{4+t}{te^{t}}$
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:
$...=\dfrac{e^{t}(-t^{2}-4t-4)}{t^{2}e^{2t}}=\dfrac{-t^{2}-4t-4}{t^{2}e^{t}}=-\dfrac{t^{2}+4t+4}{t^{2}e^{t}}$
