## Calculus with Applications (10th Edition)

$\displaystyle \frac{(2t^{3}+t^{2})e^{2t}+(2t-t^{2})e^{5t}}{(t+e^{3t})^{2}}$
$\displaystyle \frac{d}{dx}(e^{g(x)})=e^{g(x)}g^{\prime}(x)\quad (1)$ --- $y=\displaystyle \frac{t^{2}e^{2t}}{t+e^{3t}}=\frac{u(t)}{v(t)}\qquad$, we apply the quotient rule: $[\displaystyle \frac{u(t)}{v(t)}]^{\prime}=\frac{v(t)\cdot u^{\prime}(t)-u(t)\cdot v^{\prime}(t)}{[v(t)]^{2}}$ where $u^{\prime}(t)=[t^{2}e^{2t}]^{\prime}$ = product rule =$[t^{2}]^{\prime}e^{2t}+t^{2}[e^{2t}]^{\prime}$ = ... apply (1)... $=2te^{2t}+t^{2}2e^{2t}$ and $v^{\prime}(t)$= ... apply (1)... = $1+3e^{3t}$ $\displaystyle \frac{dy}{dt}=\frac{(t+e^{3t})(2te^{2t}+t^{2}2e^{2t})-t^{2}e^{2t}(1+3e^{3t})}{(t+e^{3t})^{2}}$ ... simplify... $=\displaystyle \frac{(t+e^{3t})(2te^{2t}+2t^{2}e^{2t})-t^{2}e^{2t}(1+3e^{3t})}{(t+e^{3t})^{2}}$ $=\displaystyle \frac{(2t^{2}e^{2t}+2t^{3}e^{2t}+2te^{5t}+2t^{2}e^{5t})-(t^{2}e^{2t}+3t^{2}e^{5t})}{(t+e^{3t})^{2}}$ $=\displaystyle \frac{t^{2}e^{2t}+2t^{3}e^{2t}+2te^{5t}-t^{2}e^{5t}}{(t+e^{3t})^{2}}$ $=\displaystyle \frac{(2t^{3}+t^{2})e^{2t}+(2t-t^{2})e^{5t}}{(t+e^{3t})^{2}}$