y = u^2
u = (1 - sqrt(3x - 1))
The chain rule states dy/dx = dy/(du) xx (du)/dx.
We must therefore differentiate both functions.
y' = 2u
u is a little more complicated: we will have to apply the chain rule to the radical.
The derivative of any constant c is always 0, so we don't have to worry about the 1, thankfully.
u = -v^(1/2)
v = 3x - 1
u' = -1/(2v^(1/2))
v' = 3
dy/dx = dy/(du) xx (du)/dx
dy/dx = 3 xx -1/(2v^(1/2))
dy/dx = -3/(2v^(1/2))
Now that we know the derivative of u, we can use the chain rule again to determine the derivative of f(x).
f'(x) = -3/(2v^(1/2)) xx 2u
f'(x) = -3/(2(3x - 1)^(1/2)) xx 2(1 - sqrt(3x - 1))
f'(x) = -3/(2(3x - 1)^(1/2)) xx 2 - 2sqrt(3x - 1)
f'(x) = (-3(2 - 2sqrt(3x - 1)))/(2(3x - 1)^(1/2)
f'(x) = (-6 + 6sqrt(3x- 1))/(2sqrt(3x - 1))
f'(x) = (-6(1 - 1sqrt(3x - 1)))/(2sqrt(3x - 1)
f'(x) = (-3(1 - 1sqrt(3x - 1)))/(sqrt(3x - 1)
f'(x) = (-3 + 3sqrt(3x- 1))/(sqrt(3x - 1)
f'(x) = (3sqrt(3x - 1) - 3)/(sqrt(3x - 1)
It's long, but it works!
Hopefully you understand this complicated problem better now!