# How do you differentiate y=(sinx/(1+cosx))^2?

May 5, 2017

This can be simplified to

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

$y = \frac{1 - {\cos}^{2} x}{1 + \cos x} ^ 2$

$y = \frac{\left(1 + \cos x\right) \left(1 - \cos x\right)}{1 + \cos x} ^ 2$

$y = \frac{1 - \cos x}{1 + \cos x}$

Now use the quotient rule.

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

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

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

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

Hopefully this helps!