Answer
$y'=-\dfrac{2\cos x}{(1+\sin x)^{2}}$
Work Step by Step
$y=\dfrac{1-\sin x}{1+\sin x}$
Start the differentiation process by using the quotient rule:
$y'=\dfrac{(1+\sin x)(1-\sin x)'-(1-\sin x)(1+\sin x)'}{(1+\sin x)^{2}}=...$
Evaluate the derivatives indicated in the numerator:
$...=\dfrac{-(1+\sin x)(\cos x)-(1-\sin x)(\cos x)}{(1+\sin x)^{2}}=...$
Simplify:
$...=\dfrac{-\cos x-\sin x\cos x-\cos x+\sin x\cos x}{(1+\sin x)^{2}}=...$
$...=\dfrac{-2\cos x}{(1+\sin x)^{2}}=-\dfrac{2\cos x}{(1+\sin x)^{2}}$
