Answer
$y'=\dfrac{x+\sin x}{1+\cos x}$
Work Step by Step
$y=\dfrac{x\sin x}{1+\cos x}$
Start the differentiation process by using the quotient rule:
$y'=\dfrac{(1+\cos x)(x\sin x)'-(x\sin x)(1+\cos x)'}{(1+\cos x)^{2}}=...$
Evaluate the derivatives indicated in the numerator. Use the product rule to evaluate $(x\sin x)'$:
$...=\dfrac{(1+\cos x)[x(\sin x)'+(x)'\sin x]-(x\sin x)(-\sin x)}{(1+\cos x)^{2}}=...$
Evaluate the derivatives indicated in the numerator:
$...=\dfrac{(1+\cos x)(x\cos x+\sin x)+x\sin^{2}x}{(1+\cos x)^{2}}=...$
Evaluate the operations indicated in the numerator and simplify:
$...=\dfrac{x\cos x+\sin x+x\cos^{2}x+\sin x\cos x+x\sin^{2}x}{(1+\cos x)^{2}}=...$
$...=\dfrac{x(\sin^{2}x+\cos^{2}x+\cos x)+\sin x(1+\cos x)}{(1+\cos x)^{2}}=...$
$...=\dfrac{x(1+\cos x)+\sin x(1+\cos x)}{(1+\cos x)^{2}}=...$
$...=\dfrac{(1+\cos x)(x+\sin x)}{(1+\cos x)^{2}}=\dfrac{x+\sin x}{1+\cos x}$
