I am having a problem with these two questions
1) f(t) =(3t-2) sin2t
2) f(theta) = 2cos(3theta-1)/3 theta
Help would be appreciated
$\displaystyle f(t) = (3t - 2) * sin (2t)$
Using the product and chain rules:
$\displaystyle f'(t) = (3t - 2) * 2 cos (2t) + sin (2t) * 3$
$\displaystyle f'(t) = 2*(3t - 2) cos (2t) + 3 sin (2t)$
$\displaystyle f(\theta) = \frac{2 cos (3\theta -1)}{3\theta}$
Using the chain rule and quotient rule:
$\displaystyle f'(\theta) = \frac{3\theta * -6 sin (3\theta-1) - 6 cos (3\theta-1)}{(3\theta)^2}$
$\displaystyle f'(\theta) = \frac{-18\theta * sin (3\theta-1) - 6 cos (3\theta-1)}{9\theta^2}$
$\displaystyle f'(\theta) = \frac{-6\theta * sin (3\theta-1) - 2 cos (3\theta-1)}{3\theta^2}$