Answer
$$y'=\frac{\sin \left(t\right)+t\cos \left(t\right)+t^2\cos \left(t\right)}{\left(1+t\right)^2}$$
Work Step by Step
Given
$$y=\frac{t\sin t}{ 1+t} $$
Since
\begin{align*}
y'&=\frac{\frac{d}{dt}\left(t\sin \left(t\right)\right)\left(1+t\right)-\frac{d}{dt}\left(1+t\right)t\sin \left(t\right)}{\left(1+t\right)^2}\\
&=\frac{\left(\sin \left(t\right)+t\cos \left(t\right)\right)\left(1+t\right)-1\cdot \:t\sin \left(t\right)}{\left(1+t\right)^2}\\
&=\frac{\sin \left(t\right)+t\cos \left(t\right)+t^2\cos \left(t\right)}{\left(1+t\right)^2}
\end{align*}
