Calculus 10th Edition

$f'(t)=(2\ln3-\frac{1}{t})\frac{3^{2t}}{t}$
$f(t)=y=\frac{3^{2t}}{t}$ $\ln y=\ln(\frac{3^{2t}}{t})$ $=\ln3^{2t}-\ln$ $t$ $\ln y=(2t)\ln3-\ln$ $t$ $\frac{1}{y}dy=(2\ln3-\frac{1}{t})dx$ $\frac{dy}{dx}=(2\ln3-\frac{1}{t})y$ $f'(t)=(2\ln3-\frac{1}{t})\frac{3^{2t}}{t}$