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

Nov 29, 2016

Use the quotient rule:

If we let $\cot \frac{x}{1 - \sin x} = f \frac{x}{g} \left(x\right)$

Then, to find the derivative using the quotient rule would generally look like:

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

Therefore, $y '$ would look like:

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

Find the derivatives and simplify.