(1+cosx) / (1-cosx)
By using the division rule of differentiation:
d/dx(f(x)/g(x)) = f'(x)*g(x) - g'(x)*f(x)/(g(x))^2
So, the derivative of 1 + cos(x) / 1 - cos(x) looks like:
[d/dx(1+cos(x)) * (1-cos(x)) - d/dx(1 - cos(x)) * (1 + cos(x))] / (1-cos(x))^2
which simplifies to:
[-sin(x)*(1 - cos(x)) - sin(x)*(1 + cos(x)) ] / (1-cos(x))^2
which further simplifies to:
-sin(x) +sin(x)cos(x) - sin(x) - sin(x)cos(x) / (1 - cos(x))^2
which further simplifies to:
-2sin(x) / (1 - cos(x))^2
I'm not sure if this can be further simplified. Perhaps you could expand the denominator and do some additional canceling or condensing through the use of trigonometric identities.
I hope this is helpful.