Answer
$\displaystyle \frac{(1-t)e^{3t}-4e^{2t}+(1+t)e^{t}}{(e^{2t}+1)^{2}}$
Work Step by Step
$y=\displaystyle \frac{te^{t}+2}{e^{2t}+1}=\frac{u(t)}{v(t)}$
$\begin{array}{ll}
& \text{...Apply quotient rule, } \left[\dfrac{u(x)}{v(x)}\right]^{\prime} =\dfrac{v(x)\cdot u^{\prime}(x)-u(x)\cdot v^{\prime}(x)}{[v(x)]^{2}}.\\\\
\dfrac{dy}{dt} & =\dfrac{(e^{2t}+1)(te^{t}+2)^{\prime}-(te^{t}+2)(e^{2t}+1)^{\prime}}{(e^{2t}+1)^{2}}\\
& \text{...apply product rule, }\ \ [te^{t}]'=te^{t}+e^{t}\cdot 1\\\\
& \text{... and } \quad[e^{2t}]'=e^{2t}\cdot2 \\
& \\\\
& =\dfrac{(e^{2t}+1)(te^{t}+e^{t}\cdot 1)-(te^{t}+2)(2e^{2t})}{(e^{2t}+1)^{2}} \\\\
& =\dfrac{(e^{2t}+1)(te^{t}+e^{t})-(te^{t}+2)(2e^{2t})}{(e^{2t}+1)^{2}} \\\\
& =\dfrac{te^{3t}+e^{3t}+te^{t}+e^{t}-2te^{3t}-4e^{2t}}{(e^{2t}+1)^{2}} \\\\
& =\dfrac{-te^{3t}+e^{3t}+te^{t}+e^{t}-4e^{2t}}{(e^{2t}+1)^{2}} & \\\\
& =\dfrac{(1-t)e^{3t}-4e^{2t}+(1+t)e^{t}}{(e^{2t}+1)^{2}}
\end{array}$
