Derivatives of trigonometric functions: $y= \frac{x \sin(x)}{1+\cos(x)}$ > I'm trying to find the derivative of: > > $$y= \frac{x \sin(x)}{1+\cos(x)}$$ I've tried but I can't achieve the simplified form - Here's my try- $$y' = \left(\frac{x \sin(x)}{1+\cos(x)}\right)'$$ $$y' = \frac{x\sin^2(x) + (\cos(x)+1 )(\sin(x)+x\cos(x))}{(\cos(x)+1)^2}$$ I'm pretty sure the above is correct that is why I didn't show the steps in between ... but I can't simplify it until - $$\frac{x+\sin(x)}{1+\cos(x)}$$ Which concept or formula am I missing out from in order to simplify it further? Or what should I do next? Thanks!

Note:

$$\cos^2 x + \sin^2 x = 1$$

Therefore, the numerator becomes:

$$x\sin^2x + x\cos^2x + \sin x + \cos x \sin x + x\cos x$$

$$= x(\cos^2x + \sin^2x) +\sin(x)(\cos x + 1) + x \cos x$$

$$= x(1 + \cos x) + \sin x(\cos x + 1) $$

$$= (\cos x + 1)(x + \sin x)$$

Hence, the final result becomes:

$$y' = \frac{(\cos x + 1)(x + \sin x)}{(\cos x +1)^2} = \frac{x+\sin x}{1 + \cos x}$$