Answer
$y'=(3x-2)(3x^2-4x+6)^{-1/2}$
Work Step by Step
Find y and u:
$y=u^{1/2}$
$u=3x^2-4x+6$
First, find the derivative of y and u. Don't forget to apply chain rule for y':
$y'=1/2u^{-1/2}$
$u'=6x-4$
Then plug in u and u' into y':
$y'=1/2(3x^2-4x+6)^{-1/2}\times(6x-4)=(3x-2)(3x^2-4x+6)^{-1/2}$
