用定义法求导数问题y=根号下x在x=1处的导数

定义法
y = √x
y'|(x = x0) = lim(△x→0) △y/△x = lim(x→x0) (y-y0)/(x -x0)
对于 x0 = 1
y'|(x=1)
= lim(x→1) (√x - √1)/(x - 1)
= lim(x→1) (√x -1)/[(√x + 1)(√x -1)]
= lim(x→1) 1/(√x + 1)
= 1/(√1 + 1)
= 1/2
公式法
y = √x = x^(1/2)
y' = (1/2) * x^(1/2 -1) = (1/2)*x^(-1/2) = 1/(2√x)
在 x = 1 处
y'(x=1) = 1/(2√1) = 1/2