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