Question

用导数的定义计算y=1/x 求过程

Answer 1

设y=f(x)
1
f(x+△)=1/(x+△)
f(x)=1/x
f(x+△)-f(x)=1/(x+△)-1/x=(x-x-△)/x(x+△)=-△/x(x+△)
f'(x)=lim△->0[f(x+△)-f(x)]/(x+△-x)=lim△->0[-△/x(x+△)]/△=lim△->0-1/x(x+△)=-1/x^2

2
f(x+△)=(x+△+2)^3
f(x)=(x+2)^3
f(x+△)-f(x)=(x+△+2)^3-(x+2)^3
=[x+△+2-x-2][(x+△+2)^2+(x+△+2)(x+2)+(x+2)^2]
=△[(x+△+2)^2+(x+△+2)(x+2)+(x+2)^2]

f'(x)=lim△->0[f(x+△)-f(x)]/(x+△+2-x-2)
=lim△->0 △[(x+△+2)^2+(x+△+2)(x+2)+(x+2)^2] /△
=lim△->0 [(x+△+2)^2+(x+△+2)(x+2)+(x+2)^2]
=(x+2)^2+(x+2)^2+(x+2)^2
=3(x+2)^2

Answer 2

Δy=1/(x+Δx)-1/x=-Δx/[x(x+Δx)]

Δy/Δx=-1/[x²+xΔ]

y'=limΔy/Δx=lim1/[x²+xΔ]=-1/x²
Δx→0 Δx→0