## Calculus 10th Edition

$f'(t)=-\dfrac{5}{(t-1)^2},f'(0)=-5$.
Using the quotient rule: $f'(t)=(\frac{u(t)}{v(t)})'=\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$. $u(t)=3t+2 ;u'(t)=3$. $v(t)=t-1 ;v'(t)=1$. $f'(t)=\dfrac{3(t-1)-1(3t+2)}{(t-1)^2}=-\dfrac{5}{(t-1)^2}$. $f'(0)=-\dfrac{5}{(0-1)^2}=-5.$ A graphing utility was used to verify this result.