怎样证明乘法求导法则
答案:2 悬赏:50
解决时间 2021-03-07 10:17
- 提问者网友：醉人眸
- 2021-03-06 23:16
怎样证明乘法求导法则
最佳答案
- 二级知识专家网友：都不是誰的誰
- 2021-03-07 00:21
(f(x)g(x))'
=lim(h→0)[f(x+h)g(x+h)-f(x)g(x)]/h
=lim(h→0)[f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)]/h
=lim(h→0)g(x+h)*[f(x+h)-f(x)]/h+f(x)*[g(x+h)-g(x)]/h
=g(x)f'(x)+f(x)g'(x)
(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
=lim(h→0)[f(x+h)g(x+h)-f(x)g(x)]/h
=lim(h→0)[f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)]/h
=lim(h→0)g(x+h)*[f(x+h)-f(x)]/h+f(x)*[g(x+h)-g(x)]/h
=g(x)f'(x)+f(x)g'(x)
(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
全部回答
- 1楼网友：一只傻青衣
- 2021-03-07 00:38
用定义证明
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