y=√x
dy/dx=lim(h→0) 1/h*[√(x+h)-√x]
=lim(h→0) 1/h*[√(x+h)-√x][√(x+h)+√x]/[√(x+h)+√x],分子有理化
=lim(h→0) 1/h*(x+h-x)/[√(x+h)+√x]
=lim(h→0) 1/h*h/[√(x+h)+√x]
=lim(h→0) 1/[√(x+h)+√x]
=1/[√(x+0)+√x]
=1/(√x+√x)
=1/(2√x)