关于连乘积函数高阶求导公式的教学探究

Teaching Inquiry on the Higher Derivative Formula of the Continuous Product Function

摘要: 两个函数相乘的一阶导数及高阶导数的求导法则在多数分析学教材中均已详细的给具体公式和推导过程。关于多个函数(三个及以上)之积的求导法则，尤其是高阶导数的计算公式则可以通过采取引导和启发的教学方式，同时激励学生利用所学其它课程的知识去尝试总结归纳出多函数之积的高阶求导公式，进而培养学生发散思维探究新知的能力。

Abstract: The first derivative and the higher derivative of the two functions have been detailed in most analytical textbooks. About the product of multiple functions (three or more) of the guide law, especially the higher derivative calculation formula can be guided and inspired teaching methods, motivating students to use the knowledge of other courses to try to sum up the product of the functions of higher order guide formula, and cultivate the guidance of students’ divergent thinking ability to explore new knowledge.

1. 引言

本数学分析的教学目标是培养高素质理论性加应用性的混合型人才。如何能有效的培养数学与应用数学专业大学生的发散性思维和自主探究能力，是我们在日常教学工作必须重视也必然要面对的基本问题。通过深入挖掘教材中的重要基础知识并探索更具一般性的结论有助于培养学生探究新知的能力。

在教学中我们让学生对比二项式定理与两个函数之积的莱布尼茨n阶导数公式的形式与内容，提炼出共性并让学生回顾组合数学中多项式定理的内容，进而去大胆猜测三个函数之积的莱布尼茨n阶导数公式的具体形式。通过调动学生的好奇心引导学生自己动手寻找答案。

文献中关于连乘积函数(多函数之积)求导运算方面的教学探析已有一些研究结果。如杨雄在文献 [1] 中首先对连乘积函数的一阶求导法则进行了非定义法的证明，其次给出了三函数之积的二阶导数公式的两种形式。王鹏飞在文献 [2] 中主要通过对复合函数求导法则复习总结引导学生对两个函数之积的求导法则进行了再探究，同时利用具体实例探寻多函数之积的一阶求导法则。其他相关研究结果可参见文献 [3] [4] [5] [6] [7]。文献中的研究大都集中在多函数之积的一阶求导公式或两个函数、三个函数之积的二阶导数公式的教学探究及实例应用，而通过大学其它课程知识推论论证多函数(三个及以上)之积的高阶(n阶)导数公式的教学探究文献中还未涉及。

本文通过以下具体的教学引导和理论分析环节展开教学探析。

2. 两函数之积的莱布尼茨公式

首先给出中学(高中)数学课本中的二项式定理的内容：

定理1 [8] 设 $x,y\in R$， $n\in N$，则有

$\begin{array}{c}{\left(x+y\right)}^n=C_n^0{x}^n{y}^0+C_n^1{x}^{n-1}{y}^1+\cdots +C_n^k{x}^{n-k}{y}^k+\cdots +C_n^n{x}^0{y}^n\\ ={\displaystyle \underset{k=0}{\overset{n}{\sum }}{C}_n^k{x}^{n-k}{y}^k},\end{array}$

其中 $k\in N$ 且 $0\le k\le n$， $C_n^k=\frac{n!}{k!\left(n-k\right)!}$。若二项式系数 $C_n^k$ 换成大学数学中常用记号 $\left(\begin{array}{c}n\\ k\end{array}\right)$，则二项式定理也被称为二项式公式并可表示为

${\left(x+y\right)}^n={\displaystyle \underset{k=0}{\overset{n}{\sum }}\left(\begin{array}{c}n\\ k\end{array}\right)x^{n-k}{y}^k}.$

其次给出大学课程数学分析中两个一元函数相乘的n高阶导数法则，即莱布尼茨公式：

定理2 [9] 设 $u\left(x\right),v\left(x\right)$ 均为n阶可导的函数， $n\in N$，则有函数 $u\left(x\right)v\left(x\right)$ 也n阶可导且其n阶导数为

$\begin{array}{c}{\left(uv\right)}^{\left(n\right)}=C_n^0{u}^{\left(n\right)}{v}^{\left(0\right)}+C_n^1{u}^{\left(n-1\right)}{v}^{\left(1\right)}+\cdots +C_n^k{u}^{\left(n-k\right)}{v}^{\left(k\right)}+\cdots +C_n^n{u}^{\left(0\right)}{v}^{\left(n\right)}\\ ={\displaystyle \underset{k=0}{\overset{n}{\sum }}{C}_n^k{u}^{\left(n-k\right)}{v}^{\left(k\right)}},\end{array}$

其中 $u^{\left(0\right)}=u$， $v^{\left(0\right)}=v$。该型莱布尼茨公式也可称为二函数莱布尼茨公式并简记为

${\left(uv\right)}^{\left(n\right)}={\displaystyle \underset{k=0}{\overset{n}{\sum }}\left(\begin{array}{c}n\\ k\end{array}\right)u^{\left(n-k\right)}{v}^{\left(k\right)}}.$

最后让学生找出二项式公式和二函数莱布尼茨公式中的共同之处。通过观察学生会很容易的发现两式等号右边有三处共同点：

1) 项数相同均为 $n+1$ 项；

2) 系数相同均为 $\left(\begin{array}{c}n\\ k\end{array}\right)$ ；

3) 第k项的两个变量的幂数分别和两个函数的导数阶数相同均为 $n-k$ 和k。

通过上述分析，可以引导学生去思考：如果知道了三项式定理(三项式公式)的形式，是否可以猜测出三个函数相乘的莱布尼茨公式(三函数莱布尼茨公式)的内容呢？

3. 三函数之积的莱布尼茨公式

首先给出组合数学课本中三项式定理的内容：

定理3 [10] 设 $x,y,z\in R$， $n\in N$，则有

${\left(x+y+z\right)}^n={\displaystyle \sum \left(\begin{array}{c}n\\ n_1{n}_2{n}_3\end{array}\right)x^{n_1}{y}^{n_2}{z}^{n_3}},$

其中 $\left(\begin{array}{c}n\\ n_1{n}_2{n}_3\end{array}\right)=\frac{n!}{n_1!n_2!n_3!}$， $n_1+n_2+n_3=n$， $n_1,n_2,n_3\in N$，并且项数为 $\left(\begin{array}{c}n+2\\ 2\end{array}\right)$。

设 $u\left(x\right),v\left(x\right),w\left(x\right)$ 均为n阶可导的函数， $n\in N$。此时可以引导学生猜测出三个函数相乘的莱布尼茨公式为：

定理4 设 $u\left(x\right),v\left(x\right),w\left(x\right)$ 均为n阶可导的函数， $n\in N$，则有 $u\left(x\right)v\left(x\right)w\left(x\right)$ n阶可导且其n阶导数为

${\left(uvw\right)}^{\left(n\right)}={\displaystyle \sum \left(\begin{array}{c}n\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}}.$

这时提醒学生结论可以大胆猜测，但一定要小心论证之后才能称之为定理或公式。下面引导学生用数学归纳法给出证明。

证明：当 $n=1$ 时， $\left(\begin{array}{c}1\\ 100\end{array}\right)=\left(\begin{array}{c}1\\ 010\end{array}\right)=\left(\begin{array}{c}1\\ 001\end{array}\right)=1$。于是有

$\begin{array}{c}{\left(uvw\right)}^{\left(1\right)}=u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }\\ =\left(\begin{array}{c}1\\ 100\end{array}\right)u^{\prime }vw+\left(\begin{array}{c}1\\ 010\end{array}\right)u{v}^{\prime }w+\left(\begin{array}{c}1\\ 001\end{array}\right)uv{w}^{\prime }\\ ={\displaystyle \sum \left(\begin{array}{c}1\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}}.\end{array}$

当 $n=2$ 时， $\left(\begin{array}{c}2\\ 200\end{array}\right)=\left(\begin{array}{c}2\\ 020\end{array}\right)=\left(\begin{array}{c}2\\ 002\end{array}\right)=\left(\begin{array}{c}1\\ 100\end{array}\right)=\left(\begin{array}{c}1\\ 010\end{array}\right)=\left(\begin{array}{c}1\\ 001\end{array}\right)=1$。于是有

$\begin{array}{c}{\left(uvw\right)}^{\left(2\right)}={\left({\left(uvw\right)}^{\left(1\right)}\right)}^{\prime }\\ ={\left(\left(\begin{array}{c}1\\ 100\end{array}\right)u^{\prime }vw+\left(\begin{array}{c}1\\ 010\end{array}\right)u{v}^{\prime }w+\left(\begin{array}{c}1\\ 001\end{array}\right)uv{w}^{\prime }\right)}^{\prime }\\ =\left(\begin{array}{c}1\\ 100\end{array}\right)\left(u^{″}vw+u^{\prime }{v}^{\prime }w+u^{\prime }v{w}^{\prime }\right)+\left(\begin{array}{c}1\\ 010\end{array}\right)\left(u^{\prime }{v}^{\prime }w+u{v}^{″}w+u{v}^{\prime }{w}^{\prime }\right)+\left(\begin{array}{c}1\\ 001\end{array}\right)\left(u^{\prime }v{w}^{\prime }+u{v}^{\prime }{w}^{\prime }+uv{w}^{″}\right)\\ =\left(\begin{array}{c}2\\ 200\end{array}\right)u^{″}vw+\left(\begin{array}{c}2\\ 020\end{array}\right)u{v}^{″}w+\left(\begin{array}{c}2\\ 002\end{array}\right)uv{w}^{″}+\left(\begin{array}{c}2\\ 110\end{array}\right)u^{\prime }{v}^{\prime }w+\left(\begin{array}{c}2\\ 101\end{array}\right)u^{\prime }v{w}^{\prime }+\left(\begin{array}{c}2\\ 011\end{array}\right)u{v}^{\prime }{w}^{\prime }\\ ={\displaystyle \sum \left(\begin{array}{c}2\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}},\end{array}$

其中 $\left(\begin{array}{c}2\\ 110\end{array}\right)=\left(\begin{array}{c}1\\ 100\end{array}\right)+\left(\begin{array}{c}1\\ 010\end{array}\right)$， $\left(\begin{array}{c}2\\ 101\end{array}\right)=\left(\begin{array}{c}1\\ 001\end{array}\right)+\left(\begin{array}{c}1\\ 100\end{array}\right)$， $\left(\begin{array}{c}2\\ 011\end{array}\right)=\left(\begin{array}{c}1\\ 001\end{array}\right)+\left(\begin{array}{c}1\\ 010\end{array}\right)$。

假设n时三个函数相乘的莱布尼茨公式成立，则当 $n+1$ 时可以推出

$\begin{array}{c}{\left(uvw\right)}^{\left(n+1\right)}={\left({\left(uvw\right)}^{\left(n\right)}\right)}^{\prime }={\left({\displaystyle \sum \left(\begin{array}{c}n\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}}\right)}^{\prime }\\ =(\left(\begin{array}{c}n\\ n00\end{array}\right)u^{\left(n\right)}vw+\left(\begin{array}{c}n\\ \left(n-1\right)10\end{array}\right)u^{\left(n-1\right)}{v}^{\prime }w+\left(\begin{array}{c}n\\ \left(n-1\right)01\end{array}\right)u^{\left(n-1\right)}v{w}^{\prime }+\cdots \\ +\left(\begin{array}{c}n\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}+\cdots +\left(\begin{array}{c}n\\ 01\left(n-1\right)\end{array}\right)u{v}^{\prime }{w}^{\left(n-1\right)}+{\left(\begin{array}{c}n\\ 00n\end{array}\right)uv{w}^{\left(n\right)})}^{\prime }\end{array}$

其中第二个等号后面的和式是按照 $n_1{n}_2{n}_3$ 三位数绛序展开的共有 $\left(\begin{array}{c}n+2\\ 2\end{array}\right)$ 项。对此展开式的每一项分别求一阶导数可得

${\left(\left(\begin{array}{c}n\\ n00\end{array}\right)u^{\left(n\right)}vw\right)}^{\prime }=\left(\begin{array}{c}n\\ n00\end{array}\right)\left(u^{\left(n+1\right)}vw+u^{\left(n\right)}{v}^{\prime }w+u^{\left(n\right)}v{w}^{\prime }\right),$

${\left(\left(\begin{array}{c}n\\ \left(n-1\right)10\end{array}\right)u^{\left(n-1\right)}{v}^{\prime }w\right)}^{\prime }=\left(\begin{array}{c}n\\ \left(n-1\right)10\end{array}\right)\left(u^{\left(n\right)}{v}^{\prime }w+u^{\left(n-1\right)}{v}^{″}w+u^{\left(n-1\right)}v{w}^{\prime }\right),$

${\left(\left(\begin{array}{c}n\\ \left(n-1\right)01\end{array}\right)u^{\left(n-1\right)}v{w}^{\prime }\right)}^{\prime }=\left(\begin{array}{c}n\\ \left(n-1\right)01\end{array}\right)\left(u^{\left(n\right)}v{w}^{\prime }+u^{\left(n-1\right)}{v}^{\prime }{w}^{\prime }+u^{\left(n-1\right)}v{w}^{″}\right),$

$⋮$

${\left(\left(\begin{array}{c}n\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}\right)}^{\prime }=\left(\begin{array}{c}n\\ n_1{n}_2{n}_3\end{array}\right)\left(u^{\left(n_1+1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}+u^{\left(n_1\right)}{v}^{\left(n_2+1\right)}{w}^{\left(n_3\right)}+u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3+1\right)}\right),$

$⋮$

${\left(\left(\begin{array}{c}n\\ 01\left(n-1\right)\end{array}\right)u{v}^{\prime }{w}^{\left(n-1\right)}\right)}^{\prime }=\left(\begin{array}{c}n\\ 01\left(n-1\right)\end{array}\right)\left(u^{\prime }{v}^{\prime }{w}^{\left(n-1\right)}+u{v}^{″}{w}^{\left(n-1\right)}+u{v}^{\prime }{w}^{\left(n\right)}\right),$

${\left(\left(\begin{array}{c}n\\ 00n\end{array}\right)uv{w}^{\left(n\right)}\right)}^{\prime }=\left(\begin{array}{c}n\\ 01\left(n-1\right)\end{array}\right)\left(u^{\prime }v{w}^{\left(n\right)}+u{v}^{\prime }{w}^{\left(n\right)}+uv{w}^{\left(n+1\right)}\right),$

注意到， $\forall n_1,n_2,n_3\in N$， $n_1+n_2+n_3=n+1$，有杨辉-Pascal公式：

$\left(\begin{array}{c}n+1\\ n_1{n}_2{n}_2\end{array}\right)=\left(\begin{array}{c}n\\ \left(n_1-1\right)n_2{n}_2\end{array}\right)+\left(\begin{array}{c}n\\ n_1\left(n_2-1\right)n_2\end{array}\right)+\left(\begin{array}{c}n\\ n_1{n}_2\left(n_2-1\right)\end{array}\right),$

其中如果等号右边某一项中 $\left(n_j-1\right)<0$， $1\le j\le 3$，则此项不存在。例如：

$\left(\begin{array}{c}n+1\\ \left(n+1\right)00\end{array}\right)=\left(\begin{array}{c}n\\ n00\end{array}\right),\left(\begin{array}{c}n+1\\ n10\end{array}\right)=\left(\begin{array}{c}n\\ \left(n-1\right)10\end{array}\right)+\left(\begin{array}{c}n\\ n00\end{array}\right).$

于是将上面 $\left(\begin{array}{c}n+2\\ 2\end{array}\right)$ 个等式等号两边分别相加，并利用杨辉-Pascal公式合并同类项可得三函数之积的 $n+1$ 阶导数公式为

$\begin{array}{c}{\left(uvw\right)}^{\left(n+1\right)}={\left({\left(uvw\right)}^{\left(n\right)}\right)}^{\prime }={\left({\displaystyle \sum \left(\begin{array}{c}n\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}}\right)}^{\prime }\\ =\left(\begin{array}{c}n+1\\ \left(n+1\right)00\end{array}\right)u^{\left(n+1\right)}vw+\left(\begin{array}{c}n+1\\ n10\end{array}\right)u^{\left(n\right)}{v}^{\prime }w+\left(\begin{array}{c}n+1\\ n01\end{array}\right)u^{\left(n\right)}v{w}^{\prime }+\cdots \\ +\left(\begin{array}{c}n+1\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}+\cdots +\left(\begin{array}{c}n+1\\ 01n\end{array}\right)u{v}^{\prime }{w}^{\left(n\right)}+\left(\begin{array}{c}n+1\\ 00\left(n+1\right)\end{array}\right)uv{w}^{\left(n+1\right)}\\ ={\displaystyle \underset{n_1+n_2+n_3=n+1}{\sum }\left(\begin{array}{c}n+1\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}}.\end{array}$

综上定理4得证。

例1 求 ${\left(x{\text{e}}^{2x}\sin x\right)}^{\left(5\right)}$。

解 令 $u\left(x\right)=x$， $v\left(x\right)={\text{e}}^{2x}$， $w\left(x\right)=\sin x$。由于 $u^{\left(k\right)}=x^{\left(k\right)}=0$， $k\ge 2$，并且根据三函数莱布尼茨公式，5阶求导展开式共有 $\left(\begin{array}{c}5+2\\ 2\end{array}\right)=21$ 项，则可知展开式的前10项均为0。又知 $u^{\left(k\right)}={\left({\text{e}}^{2x}\right)}^{\left(k\right)}=2^k{\text{e}}^{2x}$， $k=0,1,\cdots ,5$， $w^{\left(k\right)}={\left(\sin x\right)}^{\left(k\right)}=\sin \left(x+k\cdot \frac{\pi }{2}\right)$， $k=0,1,\cdots ,5$。于是可得

$\begin{array}{c}{\left(x{\text{e}}^{2x}\sin x\right)}^{\left(5\right)}={\left(uvw\right)}^{\left(5\right)}={\displaystyle \underset{n_1+n_2+n_3=5}{\sum }\left(\begin{array}{c}n\\ n_1{n}_2{n}_3\end{array}\right)u^{\left(n_1\right)}{v}^{\left(n_2\right)}{w}^{\left(n_3\right)}}\\ =\left(\begin{array}{c}5\\ 140\end{array}\right)u^{\prime }{v}^{\left(4\right)}w+\left(\begin{array}{c}5\\ 131\end{array}\right)u^{\prime }{v}^{\left(3\right)}{w}^{\prime }+\left(\begin{array}{c}5\\ 122\end{array}\right)u^{\prime }{v}^{″}{w}^{″}+\left(\begin{array}{c}5\\ 113\end{array}\right)u^{\prime }{v}^{\prime }{w}^{‴}\\ +\left(\begin{array}{c}5\\ 104\end{array}\right)u^{\prime }v{w}^{\left(4\right)}+\left(\begin{array}{c}5\\ 050\end{array}\right)u{v}^{\left(5\right)}w+\left(\begin{array}{c}5\\ 041\end{array}\right)u{v}^{\left(4\right)}{w}^{\prime }+\left(\begin{array}{c}5\\ 032\end{array}\right)u{v}^{‴}{w}^{″}\\ +\left(\begin{array}{c}5\\ 023\end{array}\right)u{v}^{″}{w}^{‴}+\left(\begin{array}{c}5\\ 014\end{array}\right)u{v}^{\prime }{w}^{\left(4\right)}+\left(\begin{array}{c}5\\ 005\end{array}\right)uv{w}^{\left(5\right)}\\ ={\text{e}}^{2x}\left(-38x\sin x+41x\cos x-35\sin x+120\cos x\right).\end{array}$

4. 多函数之积的莱布尼茨公式

首先给出组合数学课本中多项式定理的内容：

定理5 [10] 设 $x_1,x_2,\cdots ,x_m\in R$， $m,n\in N$，则有

${\left(x_1+x_2+\cdots +x_m\right)}^n={\displaystyle \sum \left(\begin{array}{c}n\\ n_1{n}_2\cdots n_m\end{array}\right)x_1^{n_1}{x}_2^{n_2}\cdots x_m^{n_m}},$

其中 $\left(\begin{array}{c}n\\ n_1{n}_2\cdots n_m\end{array}\right)=\frac{n!}{n_1!n_2!\cdots n_m!}$， $n_1+n_2+\cdots +n_m=n$， $n_1,n_2,\cdots ,n_m\in N$，并且项数为 $\left(\begin{array}{c}n+m-1\\ m-1\end{array}\right)$。

接下来把时间留给学生，让学生按照上述方法自己去推理论证更一般的多函数之积的莱布尼茨公式。

5. 结论与展望

通过已经掌握的课本知识去类比、归纳、推理、论证更一般的结论，有助于培养学生发散思维、创新思维的能力，也是提升学生专业技能的有效方法。关于多函数之积的极限运算、积分运算等方面的教学探究将作为今后的研究重点。

基金项目

洛阳师范学院一流本科课程《数学分析》建设项目，河南省及洛阳师范学院青年骨干教师培养计划项目(2019GGJS202, 2018XJGGJS-10)，河南省自然科学基金项目(222300420503)，河南省高校重点科研项目(22A110015)。