如何证明函数商的求导法则,即（u/v)'=(u'v-uv')/v^2越请细越好?

思路：
1/v=v的负一次方
再用乘法求导法则。

设z = u / v
那么 u = z * v
求导得 u' = z' v + z v'
所以 z' = (u' - z v') / v
将z = u / v
代入得 z' = (u' v - u v' ) / v^2

令k=1/v
(uk)' = u'k+uk'
=u' * 1/v + u * (1/v)'
= u'*1/v - u* 1/(v^2)
=(u'v-uv')/v^2