如何理解[u(ax+b)]的n阶导数=a的n次方u的n阶导数（ax+b)?

题目

如何理解[u(ax+b)]的n阶导数=a的n次方u的n阶导数（ax+b)?

答案

这是复合函数的导数：令t=ax+b
则：[u(ax+b)]'=u'(t)·(ax+b)'=au'(t)
[u(ax+b)]''=[au'(t)]'=au''(t)·(ax+b)'=a²u''(t)
[u(ax+b)]'''=[a²u''(t)]'=a²u'''(t)·(ax+b)'=a³u'''(t)
…………………………
所以：[u(ax+b)]的n阶导数=(a^n)u(t)n阶导数.