`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