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

Mar 17, 2018

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

#### Explanation:

The Quotient Rule for derivatives states:

If $f \left(x\right) = g \frac{x}{h \left(x\right)}$

then

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

So, the first step is to identify $g \left(x\right)$ and $h \left(x\right)$ in your original function:

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

so

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

$h \left(x\right) = 1 - \cos \left(x\right)$

We then differentiate the two components of $f \left(x\right)$ with respect to $x$:

$g ' \left(x\right) = \cos \left(x\right)$

$h ' \left(x\right) = - \left(- \sin \left(x\right)\right) = \sin \left(x\right)$

Let's combine all of the pieces, based on the Quotient Rule, and then simplify:

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

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

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

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

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

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

f'(x) = cancel(cos(x) - 1)/(cancel((cos(x) - 1))*(cos(x) - 1)

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