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