看的不是很清楚,给三个可能性
y=(√2)x+1
y'=lim[h→0] [√2(x+h)+1-√2x-1]/h
=lim[h→0] (√2x+√2h-√2x)/h
=lim[h→0] √2h/h
=lim[h→0] √2
=√2
y=√(2x)+1
y'=lim[h→0] {√[2(x+h)]+1-√(2x)-1}/h
=lim[h→0] [√(2x+2h)-√(2x)]/h
=lim[h→0] 1/h*[√(2x+2h)-√(2x)][√(2x+2h)+√(2x)]/[√(2x+2h)+√(2x)]
=lim[h→0] 1/h*(2x+2h-2x)/[√(2x+2h)+√(2x)]
=lim[h→0] 1/h*2h/[√(2x+2h)+√(2x)]
=lim[h→0] 2/[√(2x+2h)+√(2x)]
=2/[√(2x)+√(2x)]
=1/√(2x)
y=√(2x+1)
y'=lim[h→0] {√[2(x+h)+1]-√(2x+1)}/h
=lim[h→0] [√(2x+2h+1)-√(2x+1)]/h
=lim[h→0] 1/h*[√(2x+2h+1)-√(2x+1)][√(2x+2h+1)+√(2x+1)]/[√(2x+2h+1)+√(2x+1)]
=lim[h→0] 1/h*(2x+2h+1-2x-1)/[√(2x+2h+1)+√(2x+1)]
=lim[h→0] 1/h*2h/[√(2x+2h+1)+√(2x+1)]
=lim[h→0] 2/[√(2x+2h+1)+√(2x+1)]
=2/[√(2x+1)+√(2x+1)]
=1/√(2x+1)