## Calculus: Early Transcendentals (2nd Edition)

$\boxed{\boxed{y'=-\dfrac{1}{\sin x+1}}}$
$y=\dfrac{\cos x}{\sin x+1}$ 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: $...=-\dfrac{\sin x+1}{(\sin x+1)^{2}}=\boxed{-\dfrac{1}{\sin x+1}}$