How would you differentiate this equation?

Question

1/6(2x-3)^3 -4x

Here's my working

u = 2x-3
y = 1/6u^3
dy/du = 1/2u^2
du/dx = 2
dy/dx = dy/du x du/dx
= (2x-3)^2 -4

Is this correct? Thanks for checking!

Answer 1

(3/6)(2x-3)^2 (2) - 4

(2x-3)^2 - 4 so yes

4x^2 -12 x + 9 - 4

4 x^2 - 12 x + 5