`y= f(u) = 6u-9`

`u= g(x)`

`dy/dx = dy/(du) * (du)/(dt) `

`==> (dy)/(du) = y' = f'(u)= 6`

`==> (du)/(dx)= u' = g'(x)= (1/2)4x^3= 2x^3`

`==> (dy)/(dx) = 6 (2x^3)= 12x^3`

`==> (fog)'(x)= f'(g(x))*g'(x)`

`==> fog(x)= f(g(x))= f((1/2)x^4) = 6((1/2)x^4)-9= 3x^4 -9`

`==> (fog)'(x)= f'(g(x))* g'(x)= u' * ((1/2)x^4)'`

`==> (fog)'(x)= 6 (2x^3)= 12x^3`

We notice that (fog)'(x)= f'(g(x))* g'(x) = dy/du * du/dx

The question probably asking you to prove the derivative of fog(x) or use different methods to find the derivative.

You have 2 options:

1. Use the formula (fog)'(x)= f'(g(x)) * g'(x)

2. To find dy/dx = dy/du * du/dx