要进行二项式展开
(x+Δx)^u=C(0,u)x^(u)Δx^0+C(1,u)x^(u-1)Δx^1+C(2,u)x^(u-2)Δx^2+C(3,u)x^(u-3)Δx^3+…+C(u,u)x^(0)Δx^u
∴
(x^u)'
=lim(Δx→0)[(x+Δx)^u-x^u]/Δx
=lim(Δx→0)[C(0,u)x^(u)Δx^0+C(1,u)x^(u-1)Δx^1+C(2,u)x^(u-2)Δx^2+C(3,u)x^(u-3)Δx^3+…+C(u,u)x^(0)Δx^u-x^u]/Δx
=lim(Δx→0)[C(1,u)x^(u-1)Δx^1+C(2,u)x^(u-2)Δx^2+C(3,u)x^(u-3)Δx^3+…+C(u,u)x^(0)Δx^u]/Δx
=lim(Δx→0)C(1,u)x^(u-1)+C(2,u)x^(u-2)Δx^1+C(3,u)x^(u-3)Δx^2+…+C(u,u)x^(0)Δx^(u-1)
=C(1,u)x^(u-1)
=ux^(u-1)
哪里没看懂,再补充吧……