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

Aug 19, 2015

${y}^{'} = \frac{- 2 \sin x}{1 - \cos x} ^ 2$

#### Explanation:

You can differentiate this function by using the quotient rule and the derivative of $\cos x$, which is

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

For a function that can be written as

$\textcolor{b l u e}{y = f \frac{x}{g \left(x\right)}}$, where $g \left(x\right) \ne 0$

the quotient rule allows you to find its derivative by using the formula

color(blue)(d/dx(y) = ([d/dx(f(x))] * g(x) - f(x) * d/dx(g(x)))/((g(x))^2)

$f \left(x\right) = 1 + \cos x \text{ }$ and $\text{ } g \left(x\right) = 1 - \cos x$

This means that you can write

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left(1 + \cos x\right)\right] \cdot \left(1 - \cos x\right) - \left(1 + \cos x\right) \cdot \frac{d}{\mathrm{dx}} \left(1 - \cos x\right)}{1 - \cos x} ^ 2$

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

${y}^{'} = \frac{- \sin x + \textcolor{red}{\cancel{\textcolor{b l a c k}{\sin x \cdot \cos x}}} - \sin x - \textcolor{red}{\cancel{\textcolor{b l a c k}{\sin x \cdot \cos x}}}}{1 - \cos x} ^ 2$

${y}^{'} = \textcolor{g r e e n}{\frac{- 2 \sin x}{1 - \cos x} ^ 2}$