#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!
