2.3 高阶导数

概念

若函数 $y=f(x)$ 的导数 $f^{\prime }(x)$ 在 $x$ 处可导，则称其导数 $[f^{\prime }(x){]}^{\prime }$ 为 $f^{\prime }(x)$ 的二阶导数，记作

$$f^{\prime \prime }(x)=[f^{\prime }(x){]}^{\prime }或y^{\prime \prime }或\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}或\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}f(x)$$

同理，函数 $y=f(x)$ 的 $n$ 阶导数为 $[f^{(n-1)}(x){]}^{\prime }$ 记作

$$f^{(n)}(x)=y^{(n)}或\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}或\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}f(x)$$

运算法则

$$\begin{array}{rl}(u\pm v{)}^{(n)}& =u^{(n)}\pm v^{(n)}\\ (Cu{)}^{(n)}& =C{u}^{(n)}\\ f(\theta x{)}^{(n)}& ={\theta }^n{f}^{(n)}(\theta x)\\ (uv{)}^{(n)}& =\sum _{k=0}^n{C}_n^k{u}^{(k)}{v}^{(n-k)}(\mathrm{Leibniz}\mathrm{公式})\\ (a^x{)}^{(n)}& =a^x(\ln a{)}^n\\ [\sin (ax+b){]}^{(n)}& =a^n\sin \left(ax+b+\frac{n\pi }{2}\right)\\ [\cos (ax+b){]}^{(n)}& =a^n\cos \left(ax+b+\frac{n\pi }{2}\right)\\ {\left(\frac{1}{ax+b}\right)}^{(n)}& =\frac{(-1{)}^n{a}^{n}n!}{(ax+b{)}^{n+1}}\\ [\ln (ax+b){]}^{(n)}& =\frac{(-1{)}^{n-1}{a}^n(n-1)!}{(ax+b{)}^n}\end{array}$$

反函数的高阶导数

例 1

已知 $x=g(y)$ 是 $y=f(x)$ 的反函数，$f(x)\ne 0$，试用 $f^{\prime }(x)$，$f^{\prime \prime }(x)$，$f^{\prime \prime \prime }(x)$ 表示 $g^{\prime }(y)$，$g^{\prime \prime \prime }(y)$ .

解 由反函数求导公式知

$$g^{\prime }(y)=\frac{1}{f^{\prime }(x)}$$

所以

$$\begin{array}{rl}{g}^{\prime \prime }(y)& =\frac{\mathrm{d}}{\mathrm{d}y}\left[\frac{1}{f^{\prime }(x)}\right]\\ & =\frac{\mathrm{d}\left[\frac{1}{f^{\prime }(x)}\right]}{\mathrm{d}x}\cdot \frac{\mathrm{d}x}{\mathrm{d}y}\\ & =-\frac{f^{\prime \prime }(x)}{[f^{\prime }(x){]}^2}\cdot \frac{1}{f^{\prime }(x)}\\ & =-\frac{f^{\prime \prime }(x)}{[f^{\prime }(x){]}^3}\\ g^{\prime \prime \prime }(y)& =\frac{\mathrm{d}}{\mathrm{d}y}\left\{-\frac{f^{\prime \prime }(x)}{[f^{\prime }(x){]}^2}\right\}\\ & =\frac{\mathrm{d}\left\{-\frac{f^{\prime \prime }(x)}{[f^{\prime }(x){]}^3}\right\}}{\mathrm{d}x}\cdot \frac{\mathrm{d}x}{\mathrm{d}y}\\ & =\frac{3[f^{\prime \prime }(x){]}^2-f^{\prime }(x)f^{\prime \prime \prime }(x)}{[f^{\prime }(x){]}^4}\cdot \frac{1}{f^{\prime }(x)}\\ & =\frac{3[f^{\prime \prime }(x){]}^2-f^{\prime }(x)f^{\prime \prime \prime }(x)}{[f^{\prime }(x){]}^5}\end{array}$$

莱布尼茨公式

如果函数 $u=u(x)$ 与 $v=v(x)$ 在点 $x$ 处具有 $n$ 阶导数，考虑乘积 $u(x)v(x)$ 的 $n$ 阶导数：

由

$$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$$

得出

$$\begin{array}{c}(uv{)}^{\prime \prime }=u^{\prime \prime }v+2u^{\prime }{v}^{\prime }+u{v}^{\prime \prime }\\ (uv{)}^{\prime \prime \prime }=u^{\prime \prime \prime }v+3u^{\prime \prime }{v}^{\prime }+3u^{\prime }{v}^{\prime \prime }+v^{\prime \prime \prime }\end{array}$$

由数学归纳法

$$(uv{)}^{(n)}=u^{(n)}v+n{u}^{(n-1)}{v}^{\prime }+\frac{n(n-1)}{2!}{u}^{(n-2)}{v}^{\prime \prime }+\cdots +\frac{n(n-1)\cdots (n-k+1)}{k!}{u}^{(n-k)}{v}^{(k)}+\cdots +u{v}^{(n)}$$

即

$$(uv{)}^{(n)}=\sum _{k=0}^n{\mathrm{C}}_n^k{u}^{(n-k)}{v}^{(k)}$$

上式称为莱布尼茨公式。

例题

例 2

设 $f(x)=e^x\sin x$，求 $f^{(n)}(x)$。

NOTE

最自然的想法可能是使用莱布尼兹公式：

$$\begin{array}{rl}{f}^{(n)}(x)& =\sum _{k=1}^n{\mathrm{C}}_n^k(e^x{)}^{(k-x)}(\sin x{)}^{(k)}\\ & =\sum _{k=1}^n{\mathrm{C}}_n^k{e}^x\sin \left(x+\frac{k}{2}\pi \right)\end{array}$$

但是化到这里就化不动了。此法不可行，会扣分。

导几次之后找规律：

$$\begin{array}{rl}{y}^{\prime }& =e^x(\sin x+\cos x)=\sqrt{2}\cdot e^x\sin \left(x+\frac{\pi }{4}\right)\\ y^{\prime \prime }& =\sqrt{2}\cdot e^x\left[\sin \left(x+\frac{\pi }{4}\right)+\cos \left(x+\frac{\pi }{4}\right)\right]=(\sqrt{2}{)}^2\cdot e^x\sin \left(x+\frac{2\pi }{4}\right)\end{array}$$

归纳得 $f^{(n)}(x)=(\sqrt{2}{)}^n{e}^x\sin \left(x+\frac{n\pi }{4}\right)$。

下面使用数学归纳法证明。

显然，$n=1$ 是该式成立。

现假设 $n=k$ 时该式成立，证明 $n=k+1$ 时该式也成立。

$$\begin{array}{rl}{f}^{(k)}(x)& =(\sqrt{2}{)}^k{e}^x\sin \left(x+\frac{k\pi }{4}\right)\\ f^{(k+1)}(x)& =\frac{\mathrm{d}}{\mathrm{d}x}{f}^{(k)}(x)\\ & =(\sqrt{2}{)}^k{e}^x\left[\sin \left(x+\frac{k\pi }{4}\right)+\cos \sin \left(x+\frac{k\pi }{4}\right)\right]\\ & =(\sqrt{2}{)}^{k+1}{e}^x\sin \left(x+\frac{k\pi }{4}+\frac{\pi }{4}\right)\\ & =(\sqrt{2}{)}^{k+1}{e}^x\sin \left[x+\frac{(k+1)\pi }{4}\right]\end{array}$$

则 $n=k+1$ 时该式也成立。总上，对于任意的 $n\in {\mathbf{N}}_{+}$ 均有

$$f^{(n)}(x)=(\sqrt{2}{)}^n{e}^x\sin \left(x+\frac{n\pi }{4}\right)$$

NOTE

另外，这里提供一种超纲的解法，使用欧拉公式 $e^{i\theta }=\cos \theta +i\sin \theta$。

由 $f(x)=e^x\sin x$，构造

$$g(x)=e^x{e}^{ix}=e^x(\cos x+i\sin x)=e^x\cos x+if(x)$$

所以 $f(x)$ 是 $g(x)$ 的虚部。因此，只需求 $g^{(n)}(x)$，则 $f^{(n)}(x)$ 就是其虚部。

$$g(x)=e^x{e}^i=e^{(i+1)x}\Rightarrow g^{(n)}(x)=(i+1{)}^n{e}^{(i+1)x}$$

又有

$$\begin{array}{rl}{g}^{(n)}(x)& =(i+1{)}^n{e}^{(i+1)x}\\ & =(\sqrt{2}\cdot e^{\frac{\pi }{4}i}{)}^n{e}^x{e}^{ix}\\ & =(\sqrt{2}{)}^n{e}^{\frac{n\pi }{4}i}{e}^x{e}^{ix}\\ & =(\sqrt{2}{)}^n{e}^x{e}^{i(x+\frac{n\pi }{4}i)}\\ & =(\sqrt{2}{)}^n{e}^x\left[\cos \left(x+\frac{n\pi }{4}\right)+i\sin \left(x+\frac{n\pi }{4}\right)\right]\end{array}$$

易得其虚部为 $(\sqrt{2}{)}^n{e}^x\sin \left(x+\frac{n\pi }{4}\right)$，此即 $f^{(n)}(x)$。

例 3

求 $y=x^n\ln x$ 的 $n$ 阶导数。

$$\begin{array}{rl}k& =1,\\ & y_1=x\ln x,\\ & y_1^{\prime }=\ln x+1\\ k& =2,\\ & y_2=x^2\ln x,\\ & y_2^{\prime }=2x\ln x+x=2y_1+x,\\ & y_2^{\prime \prime }=2y_1^{\prime }+x^{\prime }\\ k& =3,\\ & y_3=x^3\ln x,\\ & y_3^{\prime }=3x^2\ln x+x^2=3y_2+x^2,\\ & y_3^{\prime \prime }=3y_2^{\prime }+(x^2{)}^{\prime }\\ & y_3^{\prime \prime \prime }=3y_2^{\prime \prime }+(x^2{)}^{\prime \prime }\\ k& =4,y_4^{\prime \prime \prime \prime }=4y_3^{\prime \prime \prime }+(x^3{)}^{\prime \prime \prime }\\ ⋮& \end{array}$$

因此推测：

$$\begin{array}{rl}{y}_n^{(n)}& =n{y}_{n-1}^{(n-1)}+(x^{n-1}{)}^{(n-1)}\\ & =n{y}_{n-1}^{(n-1)}+(n-1)!\\ & =n\left[(n-1)y_{n-2}^{(n-2)}+(n-2)!\right]+(n-1)!\\ & =n(n-1)y_{n-2}^{(n-2)}+(n-2)!n+(n-1)!\\ & =n(n-1)\left[(n-2)y_{n-3}^{(n-3)}+(n-3)!\right]+(n-2)!n+(n-1)!\\ & =n(n-1)(n-2)y_{n-3}^{(n-3)}+(n-3)!n(n-1)+(n-2)!n+(n-1)!\\ & =\frac{n!}{(n-3)!}{y}_{n-3}^{(n-3)}+\frac{n!}{n-2}+\frac{n!}{n-1}+\frac{n!}{n}\\ & =\cdots \\ & =\frac{n!}{1!}{y}_1^{\prime }+\sum _{i=2}^n\frac{n!}{i}\\ & =n!\ln x+n!+\sum _{i=2}^n\frac{n!}{i}\\ & =n!\ln x+\sum _{i=1}^n\frac{n!}{i}\end{array}$$

下面使用数学归纳法证明。

首先，$n=1$ 时有

$$y=x\ln x\Rightarrow y^{\prime }=\ln x+1=1!\ln x+\sum _{i=1}^1\frac{1!}{i}$$

故 $n=1$ 满足该式子。

下面证明当 $n=k-1$ 满足该式时，$n=k$ 也满足该式。则有

$$y_{k-1}=x^{k-1}\ln x,y_{k-1}^{(k-1)}=(k-1)!\ln x+\sum _{i=1}^{k-1}\frac{(k-1)!}{i}$$

下面考虑 $y_k$：

$$\begin{array}{rl}{y}_k=x^k\ln x\Rightarrow & y_k^{\prime }=k{x}^{k-1}\ln x+x^{k-1}=k{y}_{k-1}+(x^{k-1}{)}^{\prime }\\ \Rightarrow & y_k^{(k)}=y_k^{(k-1)}\\ & =k{y}_{k-1}^{(k-1)}+(x^{k-1}{)}^{(k-1)}\\ & =k\left[(k-1)!\ln x+\sum _{i=1}^{k-1}\frac{(k-1)!}{i}\right]+(k-1)!\\ & =k!\ln x+\sum _{i=1}^{k-1}\frac{k!}{i}+(k-1)!\\ & =k!\ln x+\sum _{i=1}^k\frac{k!}{i}\end{array}$$

证毕。