How to find derivative of y= sin^4x / cosx
precalc? sounds more like calculus.
Anyway, using the quotient rule, that's
y' = ((4sin^3x*cosx)*cosx - sin^4x(-sinx))/cos^2x
or, using the product rule, you have
y = sin^3x * tanx
y' = 3sin^2x*tanx + sin^3x*sec^2x
A little algebra should convince you that they are the same