# How do you differentiate (x)/ (1-cosx) using the quotient rule?

Jan 26, 2016

$\frac{1 - \cos x - x \sin x}{1 - \cos x} ^ 2$

#### Explanation:

The quotient rule states that

$\frac{d}{\mathrm{dx}} \left[\frac{f \left(x\right)}{g \left(x\right)}\right] = \frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{g \left(x\right)} ^ 2$

Applying this to $\frac{x}{1 - \cos x}$, we see that its derivative is equal to

$\frac{\left(1 - \cos x\right) \frac{d}{\mathrm{dx}} \left[x\right] - x \frac{d}{\mathrm{dx}} \left[1 - \cos x\right]}{1 - \cos x} ^ 2$

We can find each of the internal derivatives and then plug them back in:

$\frac{d}{\mathrm{dx}} \left[x\right] = 1$

This one is a little trickier. Since $1$ is a constant, all we are really finding is the derivative of $- \cos x$, which is positive $\sin x$.

$\frac{d}{\mathrm{dx}} \left[1 - \cos x\right] = \sin x$

Plugging it all back in:

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