# Find the derivative of $x^{-4}(3-4x^{-5})$

By Leibnitz product rule,
$f'(x) =x^{-4} \large\frac{d}{dx}$$(3-4x^{-5})+(3-4x^{-5})\large\frac{d}{dx}$$(x^{-4})$
$= x^{-4} \{0 - 4 -(-5)x^{-5-1} \} + (3-4x^{-5})(-4)x^{-4-1}$
$= x^{-4}(20x^{-6})+(3-4x^{-5})(-4x^{-5})$
$= 20x^{-10}-12x^{-5}+16x^{-10}$
$= 36x^{-10}-12x^{-5}$
$= -\large\frac{12}{x^5}$$+ \large\frac{36}{x^{10}}$
answered Apr 9, 2014