搜搜考试网函数积、商的导数证明
来源：学生作业帮 编辑：搜搜考试网作业帮 分类：数学作业 时间：2024/04/29 20:38:57
函数积、商的导数证明
额,这个要用极限证明的,因为到数的定义公式就是用极限表示的.所以至少要了解极限的含义才好啊,嗯,是大学高等数学第一章的内容.证明如下：（可参考同济版高等数学上册）
(uv)'=lim[u(x+h)v(x+h)-uv]/h
=lim[u(x+h)v(x+h)+u(x+h)v-u(x+h)v-uv]/h
=limu(x+h)[v(x+h)-v(x)]/h+limv(x)[u(x+h)-u(x)]/h
=u(x)v'(x)+u'(x)v(x)
=u'v+uv'.h→0
(u/v)'=lim(u(x+h)/v(x+h)-u(x)/v(x))/h
=lim[u(x+h)/v(x+h)-u(x)/v(x+h)+u(x)/v(x+h)-u(x)/v(x)]/h
=lim[u(x+h)-u(x)/h]/v(x+h)-limu(x)lim[v(x+h)-v(x)/h]/[v(x)v(x+h)]
=u'(x)/v(x)-u(x)v'(x)/v^2(x)
=[u'(x)v(x)-u(x)v'(x)]/v^2(x)
=(u'v-uv')/v^2.h→0
(uv)'=lim[u(x+h)v(x+h)-uv]/h
=lim[u(x+h)v(x+h)+u(x+h)v-u(x+h)v-uv]/h
=limu(x+h)[v(x+h)-v(x)]/h+limv(x)[u(x+h)-u(x)]/h
=u(x)v'(x)+u'(x)v(x)
=u'v+uv'.h→0
(u/v)'=lim(u(x+h)/v(x+h)-u(x)/v(x))/h
=lim[u(x+h)/v(x+h)-u(x)/v(x+h)+u(x)/v(x+h)-u(x)/v(x)]/h
=lim[u(x+h)-u(x)/h]/v(x+h)-limu(x)lim[v(x+h)-v(x)/h]/[v(x)v(x+h)]
=u'(x)/v(x)-u(x)v'(x)/v^2(x)
=[u'(x)v(x)-u(x)v'(x)]/v^2(x)
=(u'v-uv')/v^2.h→0