y=(2x-1)(3x+1)^2　⇔　log(y)=log(2x-1) + 2*log(3x+1)
⇔　y'/y=2/(2x-1) + 2*3/(3x+1) = {2*(3x+1) + 2*3*(2x-1)}/{(2x-1)(3x+1)}
⇔　y'={2*(3x+1) + 2*3*(2x-1)}/{(2x-1)(3x+1)}*(2x-1)(3x+1)^2
⇔　y'={2*(3x+1) + 2*3*(2x-1)}*(3x+1) = 2*(9x-2)(3x+1)