y=x+1⼀x在x=x0处的导数,定义法

2025-04-10 00:52:58
推荐回答(4个)
回答1:

如图

回答2:

定义在x = a处的导数:f'(a) = lim(x-->a) [f'(x) - f'(a)]/(x - a)
y = x + 1/x
y'(x₀) = lim(x-->x₀) [(x + 1/x) - (x₀ + 1/x₀)]/(x - x₀)
= lim(x-->x₀) [(x² + 1)/x - (x₀² + 1)/x₀]/(x - x₀)
= lim(x-->x₀) [(x² + 1)x₀ - (x₀² + 1)x]/[xx₀(x - x₀)]
= lim(x-->x₀) (x₀x² + x₀ - xx₀² - x)/[xx₀(x - x₀)]
= lim(x-->x₀) [(x₀x - 1)x - (x₀x - 1)x₀]/[xx₀(x - x₀)]
= lim(x-->x₀) [(x₀x - 1)(x - x₀)]/[xx₀(x - x₀)]
= lim(x-->x₀) (x₀x - 1)/xx₀
= (x₀² - 1)/x₀²
= 1 - 1/x₀²

回答3:

解:y的导数=△y/△x=(x+△x+1/(x+△x)-x-1/x)/△x=1-1/(x+△x)*x=1-1/(x0+0)x0=1-1/x0^2

回答4:

y'=1-1/(x0^2)