Answer
(a) $dy = (cot~\theta)~d\theta$
(b) $dy = \frac{e^x}{(1-e^x)^2}~dx$
Work Step by Step
(a) $y = ln~(sin~\theta)$
$\frac{dy}{d\theta} = \frac{1}{sin~\theta}\cdot~(cos~\theta)$
$\frac{dy}{d\theta} = \frac{cos~\theta}{sin~\theta}$
$\frac{dy}{d\theta} = cot~\theta$
$dy = (cot~\theta)~d\theta$
(b) $y = \frac{e^x}{1-e^x}$
$\frac{dy}{dx} = \frac{(e^x)(1-e^x)-(e^x)(-e^x)}{(1-e^x)^2}$
$\frac{dy}{dx} = \frac{e^x-e^{2x}+e^{2x}}{(1-e^x)^2}$
$\frac{dy}{dx} = \frac{e^x}{(1-e^x)^2}$
$dy = \frac{e^x}{(1-e^x)^2}~dx$
