How do you find the derivative of #f(x) = 2/(x-1) - 1/(x+2) #?

As this is posted n the "Quotient Rule" I will use it, but is not the easiest way to deal with this particular problem, as the chain rule is much quicker, and less prone to introducing an error (see Method 2)

Method 1 - Using The Quotient Rules:

The quotient rule states that:
# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #
It will need to be applied to each term
# f(x)=2/(x-1)-1/(x+2) #

# :. f'(x) = { ((x-1)(d/dx2)-(2)(d/dx(x-1)))/(x-1)^2 } - { ((x+2)(d/dx2)-(1)(d/dx(x+2)))/(x+2)^2 } #

# :. f'(x) = { (0-(2)(1))/(x-1)^2 } - { (0-(1)(1))/(x+2)^2 } #
# :. f'(x) = { (-2)/(x-1)^2 } - { (-1)/(x+2)^2 } #
# :. f'(x) = (-2)/(x-1)^2 + 1/(x+2)^2 #

Method 2 - Using The Chain Rule instead of the Quotient Rules:

# f(x)=2/(x-1)-1/(x+2) #
# :. f(x)=2(x-1)^-1 -(x+2)^-1 #

Implicit using the chain rule gives;
# f'(x)=2(-1)(x-1)^(-1-1)(1) - (-1)(x+2)^(-1-1)(1) #
# :. f'(x)=-2(x-1)^-2 + (x+2)^-2 #
# :. f'(x)=-2/(x-1)^2 + 1/(x+2)^2 # (as above)

If you can't follow the implicit application of the chain rule,let me know and I will add explicit steps for clarification.