Answer
a. $dy = \frac{-du}{(1+3u)^2}$
b. $dy = 2\theta(sin 2\theta + \theta cos(2\theta)) d\theta$
Work Step by Step
a. $y = \frac{1+2u}{1+3u}$
$f'(u) = \frac{2(1+3u) - 3(1+2u)}{(1+3u)^2}$
$f'(u) = \frac{2+6u - 3-6u}{(1+3u)^2}$
$f'(u) = \frac{-1}{(1+3u)^2}$
$dy = f'(u)du$
$dy = \frac{-1}{(1+3u)^2}du$
$dy = \frac{-du}{(1+3u)^2}$
b. $y = \theta^2 sin 2\theta$
$f'(\theta) = 2\theta sin2\theta + 2\theta^2 cos(2\theta)$
$f'(\theta) = 2\theta(sin 2\theta + \theta cos(2\theta))$
$dy = f'(\theta)d\theta$
$dy = 2\theta(sin 2\theta + \theta cos(2\theta)) d\theta$
