## Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{t^{2}\cos t+t\cos t+\sin t}{(1+t)^{2}}$
$y=\dfrac{t\sin t}{1+t}$ Differentiate using the quotient rule: $y'=\dfrac{(1+t)(t\sin t)'-(t\sin t)(1+t)'}{(1+t)^{2}}=...$ Use the product rule to find $(t\sin t)'$: $...=\dfrac{(1+t)[(t)(\sin t)'+(\sin t)(t)']-(t\sin t)(1)}{(1+t)^{2}}=...$ $...=\dfrac{(1+t)(t\cos t+\sin t)-t\sin t}{(1+t)^{2}}=...$ Simplify: $...=\dfrac{t\cos t+\sin t+t^{2}\cos t+t\sin t-t\sin t}{(1+t)^{2}}=...$ $...=\dfrac{t^{2}\cos t+t\cos t+\sin t}{(1+t)^{2}}$