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

##### 1 Answer

#### Explanation:

You can differentiate this function by using the quotient rule and the derivative of

#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)#