Calculus: Early Transcendentals (2nd Edition)

$p'(x) = \frac{-4}{x^3} - \frac{6}{x^5} - \frac{5}{2x^6}$
First Expand Fraction Into a Sum of Three Terms $p(x) = \frac{4x^3+3x+1}{2x^5} = 2x^{-2}+\frac{3}{2}x^{-4}+\frac{1}{2}x^{-5}$ Power Rule: $p'(x) = 2(-2)x^{(-2-1)} + \frac{3}{2}(-4)x^{(-4-1)} + \frac{1}{2}(-5)x^{(-5-1)} = -4x^{-3}-6x^{-5}-\frac{5}{2}x^{-6} = \frac{-4}{x^3} - \frac{6}{x^5} - \frac{5}{2x^6}$