How would you differentiate this equation?
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!
(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