Raymond B. answered 06/10/21
Math, microeconomics or criminal justice
y = (2x+3)^3(3x-2)^4
dy/dx = y' = (2x+3)^3(4)(3x-2)^3(3) + (3x-2)^4(3)(2x+3)^2(2)
=12(2x+3)^3(3x-2)^3 + 6(3x-2)^4(2x+3)^2
derivative of a product = 1st term times derivative of the 2nd, plus 2nd times derivative of the 1st
y=(x^2+x/2 -1)^3 use the chain rule
dy/dx = y'= 3(x^2+ x/2 -1)^2(2x+ 1/2)