# How do you find the derivative of ((sinx)^2)/(1-cosx)?

Jun 17, 2016

$- \sin x$

#### Explanation:

The derivative of the quotient $\frac{u}{v}$

$d \left(\frac{u}{v}\right) = \frac{u ' v - v ' u}{v} ^ 2$

Let $u = {\left(\sin x\right)}^{2}$ and $v = 1 - \cos x$

$\frac{d {\left(\sin x\right)}^{2}}{\mathrm{dx}} = 2 \sin \left(x\right) \cdot \frac{\mathrm{ds} \in x}{\mathrm{dx}}$

$= 2 \sin x \cos x$

$\textcolor{red}{u ' = 2 \sin x \cos x}$

$\frac{d \left(1 - \cos \left(x\right)\right)}{\mathrm{dx}} = 0 - \left(- \sin x\right) = \sin x$

$\textcolor{red}{v ' = \sin x}$

Apply the derivative property on the given quotient:

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

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

Simplify by $1 - \cos x$ this leads to

$= \frac{2 \sin x \cos x - \sin x \left(1 + \cos x\right)}{1 - \cos x}$
$= \frac{2 \sin x \cos x - \sin x - \sin x \cos x}{1 - \cos x}$
$= \frac{\sin x \cos x - \sin x}{1 - \cos x}$
$= \frac{- \sin x \left(- \cos x + 1\right)}{1 - \cos x}$
$= \frac{- \sin x \left(1 - \cos x\right)}{1 - \cos x}$
Simplify by $1 - \cos x$

$= - \sin x$