导数除法法则证明
2个回答
展开全部
等号右边就有错,除法法则是两个函数相除之后的新函数的导数,而非上下导数相除
(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
(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
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询