Let's start off with the definition of the quotient rule:
#d/dx (f(x)/(g(x))) = (g(x)f'(x) - f(x)g'(x))/(g(x))^2#
Or to make remembering the rule simpler, let #f(x)# be "#high#," being the upper function and let #g(x)# be "#low#," being the lower function to give us:
#d/dx ((high)/(low)) = (low*d'(high) - (high*d'(low)))/(low)^2#
where #d'# denotes the "derivative of."
This formula can be read as "low d high minus high d low over low squared," which has a nice flow for memorization.
In the case #f(x) = (1-x)/(x^3 - 6)#, let "#low#" denote #x^3 - 6# and "#high#" denote #1-x#
Applying the quotient rule,
#d/dx ((high)/(low)) = ((x^3 - 6)*d'(1-x) - ((1-x)*d'(x^3 - 6)))/(x^3 - 6)^2#
Now switching #((high)/(low))# to #f(x)# for simplicity,
#d/dx (f(x)) = ((x^3 - 6)⋅(-1) - ((1-x)*(3x^2)))/(x^3 - 6)^2#
Knowing that by the power rule #d/dx (x^n) = nx^(n-1)#
And that the derivative of a constant is #0#.
#d/dx (f(x)) = ((-x^3 + 6) - ((3x^2-3x^3)))/(x^3 - 6)^2#
#d/dx (f(x)) = (-x^3 + 6 - 3x^2+3x^3)/(x^3 - 6)^2 = (2x^3-3x^2+6)/((x^3 - 6)^2)#
