## Calculus 8th Edition

$2e^{2t}\sec^2(1+e^{2t})$
$\frac{d}{dt}\tan(1+e^{2t})$ Use chain rule: $=\sec^2(1+e^{2t})\frac{d}{dt}(1+e^{2t})$ $=\sec^2(1+e^{2t})*e^{2t}*\frac{d}{dt}(2t)$ $=\sec^2(1+e^{2t})*e^{2t}*2$ $=2e^{2t}\sec^2(1+e^{2t})$