# Question #50154

Nov 10, 2016

Use a combination of the product and quotient rules.

Let $g \left(x\right) = x \sin x$.

$g ' \left(x\right) = 1 \times \sin x + x \times \cos x = \sin x + x \cos x$

Let the entire function be $f \left(x\right) = \frac{x \sin x}{1 + \cos x}$.

$f ' \left(x\right) = \frac{\left(\sin x + x \cos x\right) \left(1 + \cos x\right) - \left(- \sin x \times x \sin x\right)}{1 + \cos x} ^ 2$

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

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

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

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

$f ' \left(x\right) = \frac{\left(\sin x + x\right) \left(1 + \cos x\right)}{\left(1 + \cos x\right) \left(1 + \cos x\right)}$

$f ' \left(x\right) = \frac{\sin x + x}{1 + \cos x}$

Hopefully this helps!