Question

How do you differentiate `f(x)=(2x^2-5)^3/(x^3+6x)^(4/3)` using the quotient rule?

Answer 1

write f = N/D then `f' = (DN'-D'N)/D^2` With `N = (2x^2 - 5)^3` and `D = (x^3 + 6x)^(4/3)` one obtains
`N' = 3(2x^2-5)^2 4x`
`D' = 4/3(x^3 + 6x)^(1/3) (3x^2 + 6)`
so
`f' = [(x^3 + 6x)^(4/3)3(2x^2-5)^2 4x - (2x^2 - 5)^(3)(x^3 + 6x)^(1/3) (3x^2 + 6)]/(x^3 + 6x)^(8/3)`
simplifying yields
`f' = (x^3 + 6x)^(-7/3)(2x^2-5)^2 [(x^3 + 6x)12x - (2x^2 - 5)(3x^2 + 6)]`
`f' = (x^3 + 6x)^(-7/3)(2x^2-5)^2 [6x^4 + 75x^2 + 30}`