导数运算法则是怎么推出来的

2025-03-26 09:38:56
推荐回答(4个)
回答1:

求函数y=f(x)在x0处导数的步骤:
① 求函数的增量Δy=f(x0+Δx)-f(x0)
② 求平均变化率
③ 取极限,得导数。

说得具体点,就是在函数上取相近的两点,求这两点的斜率,当这两点足够近时(取极限),所得的值就是函数在该点的导数。一般求导都是直接用导数公式(靠记忆)

用极限推导,在选修2-2里

(f(x)g(x))'
=im(x+h)v(x+h)-u(x)v(x)]/h}
=lin(x+h)v(x+h)-u(x)v(x+h)]/h}+LIM{[u(x)v(x+h)-u(x)v(x)]/h}
=u(x)'v(x)+u(x)v(x)'

[f(x)/g(x)]'
=lim(Δx→0)(f(x+Δx)/g(x+Δx)-f(x)/g(x))/Δx)
=lim(Δx→0)((g(x)*f(x+Δx)-f(x)*g(x+Δx))/(g(x+Δx)*g(x)))/Δx)
=lim(Δx→0)((g(x)*f(x+Δx)/Δx-f(x)*g(x+Δx)/Δx)/(g(x+Δx)*g(x)))
=lim(Δx→0)(g(x)*f(x+Δx)/Δx-f(x)*g(x+Δx)/Δx)/lim(Δx→0)(g(x+Δx)*g(x))
=lim(Δx→0){[g(x)*f(x +Δx)-g(x)*f(x)]/Δx+[g(x)*f(x)-f(x)*g(x+Δx)]/Δx}/lim(Δx→0)(g(x+Δx)*g(x))
=(f'(x)g(x)-f(x)g'(x))/(g(x))²

回答2:

导数(Derivative)是微积分中的重要基础概念。当自变量的增量趋于零时,因变量的增量与自变量的增量之商的极限。在一个函数存在导数时,称这个函数可导或者可微分。可导的函数一定连续。不连续的函数一定不可导。导数实质上就是一个求极限的过程,导数的四则运算法则来源于极限的四则运算法则。
导数通常用于求曲线的切线斜率。
具体的推导过程,你可以看看这个参考资料,这是大一高等数学里,我想你应该能看明白的

回答3:

F(▲X)-F(▲X+X)
除以
▲X
好像是这样的

回答4:

就说说除法吧:[f(x)/g(x]"={(fx+dfx)/(gx+dgx)-f(x)/g(x}/dx=
你算算吧,不好意思,网管让我下机了,没钱充值了

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