Question

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

Answer 1

`y^' = (-2sinx)/(1-cosx)^2`

Explanation:

You can differentiate this function by using the quotient rule and the derivative of `cosx`, which is

`d/dx(cosx) = -sinx`

For a function that can be written as

`color(blue)(y = f(x)/(g(x)))`, where `g(x)!=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)`

In your case,

`f(x) = 1 + cosx" "` and `" "g(x) = 1 - cosx`

This means that you can write

`d/dx(y) = ([d/dx(1 + cosx)] * (1-cosx) - (1 + cosx) * d/dx(1-cosx))/(1-cosx)^2`

`y^' = (-sinx * (1 - cosx) - (1 + cosx) * (-(-sinx)))/(1-cosx)^2`

`y^' = (-sinx + color(red)(cancel(color(black)(sinx * cosx))) - sinx - color(red)(cancel(color(black)(sinx * cosx))))/(1-cosx)^2`

`y^' = color(green)((-2sinx)/(1-cosx)^2)`