## Calculus 8th Edition

$F'(t)=-2ln(3) sin(2t) 3^{cos2t}$
Given: $F(t)=3^{cos2t}$ $ln[F(t)]=ln(3^{cos2t})$ $\frac{d}{dt}lnF(t)=\frac{d}{dt}ln(3^{cos2t})$ $\frac{d}{dt}lnF(t)=\frac{d}{dt}[cos2t$ ln$3]$ $\frac{1}{F(t)}F'(t)=ln3 (-sin2t) 2$ $F'(t)=F(t) ln3 (-sin2t) 2$ Hence, $F'(t)=-2ln(3) sin(2t) 3^{cos2t}$