Question: (A) f(x) = (3x - 2) / 5x² _OR_ (B) f(x) = 3x - (2 / 5x²) ?
Comment: You _cannot_ be using PEMDAS (without Parentheses) in Calculus (or Algebra for that matter)… just no ! …well, within reason.
Answer: There is no question here, so I’ll just make up something obvious... Find f’(x)… both ways.
(A) f(x) = (3x - 2) / 5x² ==>
f'(x) = [ (5x²)(3) - (3x - 2)(10x) ] / (5x²)² = (15x² - 30x² + 20x) / 25x⁴ = (-15x² + 20x) / 25x⁴
= (-3x² + 4x) / 5x⁴ = (4 - 3x) / 5x³
(B) f(x) = 3x - (2/5)(1/x²) = 3x - (2/5)(x⁻²) ==>
f'(x) = 3 - (2/5)(-2)x⁻³ = 3 - (2/5)(-2)(1/x³) = 3 + 4/(5x³)
