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

Sep 4, 2016

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

#### Explanation:

We have: $y = \frac{\sin \left(x\right)}{1 - \cos \left(x\right)}$

This function can be differentiated using the "quotient rule":

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

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

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

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

Let's apply the Pythagorean identity ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$:

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

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

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

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