How to differentiate a bracket raised to a power i.e. chain rule

Lets say the equation to be differentiated takes the following format y = (ax²+bx+c)ⁿ, to find dy/dx: (1) Make u equal the contents of the bracket, u=ax²+bx+c. (2) Substitute the contents of the bracket with u, y=uⁿ. (3) Differentiate y with respect to u, dy/du=nu^(n-1). (4) Differentiate u with respect to x, du/dx=2ax+b. (5) Because dy/dx=(dy/du)(du/dx), we can derive. (6) dy/dx=(nu^(n-1))(2ax+b). (7) Finally, substitute u with ax²+bx+c, dy/dx=(n(ax²+bx+c)^(n-1))(2ax+b).