求推导微积分 d(xy)＝xd(y)+yd(x)

Question

如题所述

Answer 1

（xy）’=xy’+y两边都是对x求导，所以得到
=>d(xy)/dx=x*(dy/dx)+y
=>d(xy)=xdy +ydx

Answer 2

令u(x)，v(x)
则根据导数定义，u'(x)=lim(t->0) [u(x+t)-u(x)]/t，v'(x)=lim(t->0) [v(x+t)-v(x)]/t
(u*v)'=lim(t->0) [u(x+t)v(x+t)-u(x)v(x)]/t
=lim(t->0) [u(x+t)v(x+t)-u(x)v(x+t)+u(x)v(x+t)-u(x)v(x)]/t
=lim(t->0) v(x+t)[u(x+t)-u(x)]/t+lim(t->0) u(x)[v(x+t)-v(x)]/t
=v(x)*u'(x)+u(x)*v'(x)

Answer 3

教材上有这个推导的。