问答题
问答题
设函数u(x)，v(x)可导，利用导数定义证明[u(x)v(x)]"=u"(x)v(x)+u(x)v"(x)；
【正确答案】
【答案解析】解：
问答题
设函数u
1
(x)，u
2
(x)，…，u
n
(x)可导，f(x)=u
1
(x)u
2
(x)…u
n
(x)，写出f(x)的求导公式。
【正确答案】
【答案解析】解：由题意得
f"(x)={u 1 (x)[u 2 (x)…u n (x)]}"
=u 1 "(x)·[u 2 (x)…u n (x)]+u 1 (x)[u 2 (x)…u n (x)]"
=u 1 "(x)·u 2 (x)…u n (x)+u 1 (x){u 2 (x)·[u 3 (x)…u n (x)]}"
…
=u 1 "(x)·u 2 (x)…u n (x)+u 1 (x)·u 2 "(x)…u n (x)+…+u 1 (x)·u 2 (x)…u n "(x)。
f"(x)={u 1 (x)[u 2 (x)…u n (x)]}"
=u 1 "(x)·[u 2 (x)…u n (x)]+u 1 (x)[u 2 (x)…u n (x)]"
=u 1 "(x)·u 2 (x)…u n (x)+u 1 (x){u 2 (x)·[u 3 (x)…u n (x)]}"
…
=u 1 "(x)·u 2 (x)…u n (x)+u 1 (x)·u 2 "(x)…u n (x)+…+u 1 (x)·u 2 (x)…u n "(x)。