(x+h)^(1/3)如何解 急啊

来源：百度知道 编辑：UC知道 时间：2024/05/18 03:32:52
题目中让我先证明 a-b={a^(1/3)-b^(1/3)}{a^(2/3)+a^(1/3)b^(1/3)+b^(2/3)}
第二问让我根据这个证明来求 d/dx x^(1/3)
要求用first principles
就是这个公式limit {f(x+h)-f(x)}/h
要怎么做啊?

a-b={a^(1/3)-b^(1/3)}{a^(2/3)+a^(1/3)b^(1/3)+b^(2/3)}
这就是立方差
x^3-y^3=(x-y)(x^2+xy+y^2)
a-b={a^(1/3)-b^(1/3)}{a^(2/3)+a^(1/3)b^(1/3)+b^(2/3)}
所以a^(1/3)-b^(1/3)=(a-b)/{a^(2/3)+a^(1/3)b^(1/3)+b^(2/3)}

[(x+h)^(1/3)]^3=x+h
[x^(1/3)]^3=x

所以 {f(x+h)-f(x)}/h
=[(x+h)^(1/3)-x^(1/3)]/h
={[(x+h)-x]/[(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]}/h
={h/[(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]}/h
=1/[(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]
h→0
所以极限=1/[x^(2/3)+x^(1/3)*x^(1/3)+x^(2/3)]
=1/[3x^(2/3)]
即(1/3)x^(-2/3)