设y=根号x分之x²-x+根号x-1,求导数与微分?

设y=根号x分之x²-x+根号x-1,求导数与微分?
设y=根号x分之x²-x+根号x-1,求导数与微分?

设y=根号x分之x²-x+根号x-1,求导数与微分?
y=(x²-x)/√x+√(x-1)
=x^(3/2)-√x+√(x-1)
所以
y'=3/2*√x-1/2√x+1/2√(x-1)
=(3x-1)/(2√x)+1/2√(x-1)
dy/dx=y'
所以
dy=[(3x-1)/(2√x)+1/2√(x-1)]dx