Answer
$y'=\dfrac{1}{1-\sin x}$
Work Step by Step
$y=\dfrac{\cos x}{1-\sin x}$
Differentiate using the quotient rule:
$y'=\dfrac{(1-\sin x)(\cos x)'-(\cos x)(1-\sin x)'}{(1-\sin x)^{2}}=...$
$...=\dfrac{(1-\sin x)(-\sin x)-(\cos x)(-\cos x)}{(1-\sin x)^{2}}=...$
$...=\dfrac{-\sin x+\sin^{2}x+\cos^{2}x}{(1-\sin x)^{2}}=...$
We know $\sin^{2}x+\cos^{2}x=1$, so we substitute and simplify:
$...=\dfrac{1-\sin x}{(1-\sin x)^{2}}=\dfrac{1}{1-\sin x}$