use logarithmic differential to find dy/dx, where

$\displaystyle y = \frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}$

$\displaystyle y' = y*$________

sofar:

$\displaystyle ln y = ln\frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}$

= $\displaystyle -3ln(x-4) + 4ln(sin(x)) - 2ln(x^2+2x) + x^3$

= $\displaystyle \frac{1}{y}y' = \frac{-3}{x-4} + \frac{4}{cos(x)} -2\frac{2x+2}{x^2+2x} + 3x^2$

not sure what to do next