# What is the derivative of this function y= (1+sin x)/(1-sin x)?

Nov 5, 2016

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

#### Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $y = \frac{1 + \sin x}{1 - \sin x}$ Then

$\left\{\begin{matrix}\text{Let "u=1+sinx & => & (du)/dx=cosx \\ "And } v = 1 - \sin x & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = - \cos x\end{matrix}\right.$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \sin x\right) \left(\cos x\right) - \left(1 + \sin x\right) \left(- \cos x\right)}{1 - \sin x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos x - \sin x \cos x + \cos x + \sin x \cos x}{1 - \sin x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \cos x}{1 - \sin x} ^ 2$