Derivatives of trigonometric functions: $y= \frac{x \sin(x)}{1+\cos(x)}$

Question

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!

Answer 1

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:

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

Answer 2

Note that $x\sin^2 x = x(1 - \cos^2 x)$. So we can rewrite the numerator as $$x-x\cos^2 x + x\cos^2 x+(1+\cos x)\sin x +x\cos x = (1+\cos x)(x+\sin x)$$ so $$y'=\frac{(1+\cos x)(x+\sin x)}{(1+\cos x)(1+\cos x)} = \frac{x+\sin x}{1+\cos x}$$

Answer 3

HINT:

Using Weierstrass Substitution, $\dfrac{\sin x}{1+\cos x}=\tan\dfrac x2$

$$\dfrac{d\left(\dfrac{x\sin x}{1+\cos x}\right)}{dx}=\tan\dfrac x2+\dfrac{x\sec^2\dfrac x2}2$$

Now $\dfrac{\sec^2\dfrac x2}2=\dfrac1{2\cos^2\dfrac x2}=\dfrac1{1+\cos x}$