【数学分析笔记】第4章第5节 高阶导数和高阶微分（2）
4.5 高阶导数和高阶微分
【例4.5.3】
y
=
x
m
,
m
y=x^m,m
y=xm,m是正整数，求
y
(
n
)
y^{(n)}
y(n).
【解】
y
′
=
m
x
m
−
1
y'=mx^{m-1}
y′=mxm−1
y
′
′
=
m
(
m
−
1
)
x
m
−
2
y''=m(m-1)x^{m-2}
y′′=m(m−1)xm−2
y
′
′
′
=
m
(
m
−
1
)
(
m
−
2
)
x
m
−
3
y'''=m(m-1)(m-2)x^{m-3}
y′′′=m(m−1)(m−2)xm−3
…
y
(
m
)
=
m
(
m
−
1
)
.
.
.
2
⋅
1
x
0
=
m
!
y^{(m)}=m(m-1)...2\cdot1x^0=m!
y(m)=m(m−1)...2⋅1x0=m!
y
(
m
+
1
)
=
0
y^{(m+1)}=0
y(m+1)=0
则
(
x
m
)
(
n
)
=
{
m
(
m
−
1
)
.
.
.
(
m
−
n
+
1
)
x
m
−
n
,
n
≤
m
0
,
n
>
m
(x^m)^{(n)}=\left\{\begin{matrix} m(m-1)...(m-n+1) x^{m-n}&, n\le m \\ 0&,n>m \end{matrix}\right.
(xm)(n)={m(m−1)...(m−n+1)xm−n0,n≤m,n>m
【例4.5.4】
y
=
ln
x
y=\ln x
y=lnx，求
y
(
n
)
y^{(n)}
y(n).
【解】
y
′
=
1
x
=
x
−
1
=
1
x
y'=\frac{1}{x}=x^{-1}=\frac{1}{x}
y′=x1=x−1=x1
y
′
′
=
−
x
−
2
=
−
1
x
2
y''=-x^{-2}=-\frac{1}{x^2}
y′′=−x−2=−x21
y
′
′
′
=
(
−
1
)
(
−
2
)
x
−
3
=
2
x
3
y'''=(-1)(-2)x^{-3}=\frac{2}{x^3}
y′′′=(−1)(−2)x−3=x32
y
(
4
)
=
(
−
1
)
(
−
2
)
(
−
3
)
x
−
4
=
−
3
!
x
4
y^{(4)}=(-1)(-2)(-3)x^{-4}=-\frac{3!}{x^4}
y(4)=(−1)(−2)(−3)x−4=−x43!
…
于是
y
(
n
)
=
(
−
1
)
n
−
1
(
n
−
1
)
!
x
n
y^{(n)}=(-1)^{n-1}\frac{(n-1)!}{x^n}
y(n)=(−1)n−1xn(n−1)!
【例】
y
=
1
x
y=\frac{1}{x}
y=x1，求
y
(
n
)
y^{(n)}
y(n).
【解】
(
1
x
)
(
n
)
=
(
ln
x
)
(
n
+
1
)
=
(
−
1
)
n
n
!
x
n
+
1
(\frac{1}{x})^{(n)}=(\ln x)^{(n+1)}=(-1)^{n}\frac{n!}{x^{n+1}}
(x1)(n)=(lnx)(n+1)=(−1)nxn+1n!
4.5.1 高阶导数运算法则
【定理4.5.1】
f
(
x
)
,
g
(
x
)
f(x),g(x)
f(x),g(x)都是
n
n
n次可导，则
(
c
1
f
(
x
)
+
c
2
g
(
x
)
)
(
n
)
=
c
1
f
(
n
)
(
x
)
+
c
2
g
(
n
)
(
x
)
(c_1f(x)+c_2g(x))^{(n)}=c_1f^{(n)}(x)+c_2g^{(n)}(x)
(c1f(x)+c2g(x))(n)=c1f(n)(x)+c2g(n)(x).
推广到
n
n
n项有：
【定理4.5.2】【Leibniz（莱布尼茨）公式】
f
(
x
)
,
g
(
x
)
f(x),g(x)
f(x),g(x)都是
n
n
n次可导，则
(
f
(
x
)
⋅
g
(
x
)
)
(
n
)
=
∑
k
=
0
n
C
n
k
f
(
n
−
k
)
(
x
)
g
(
k
)
(
x
)
(f(x)\cdot g(x))^{(n)}=\sum\limits_{k=0}^{n}\mathrm{C}_{n}^{k}f^{(n-k)}(x)g^{(k)}(x)
(f(x)⋅g(x))(n)=k=0∑nCnkf(n−k)(x)g(k)(x)
【证】用数学归纳法
设
n
=
1
n=1
n=1，
(
f
(
x
)
⋅
g
(
x
)
)
′
=
f
′
(
x
)
g
(
x
)
+
f
(
x
)
g
′
(
x
)
=
∑
k
=
0
1
C
1
k
f
(
n
−
k
)
(
x
)
g
(
k
)
(
x
)
(f(x)\cdot g(x))'=f'(x)g(x)+f(x)g'(x)=\sum\limits_{k=0}^{1}\mathrm{C}_{1}^{k}f^{(n-k)}(x)g^{(k)}(x)
(f(x)⋅g(x))′=f′(x)g(x)+f(x)g′(x)=k=0∑1C1kf(n−k)(x)g(k)(x)
n
=
1
n=1
n=1时，莱布尼茨公式成立
设莱布尼茨公式对
n
=
m
n=m
n=m成立，即
(
f
(
x
)
g
(
x
)
)
(
m
)
=
∑
k
=
0
m
C
m
k
f
(
n
−
k
)
(
x
)
g
(
k
)
(
x
)
(f(x)g(x))^{(m)}=\sum\limits_{k=0}^{m}\mathrm{C}_{m}^{k}f^{(n-k)}(x)g^{(k)}(x)
(f(x)g(x))(m)=k=0∑mCmkf(n−k)(x)g(k)(x)
(
f
(
x
)
g
(
x
)
)
(
m
+
1
)
=
(
(
f
(
x
)
g
(
x
)
)
(
m
)
)
′
=
∑
k
=
0
m
C
m
k
(
f
(
n
−
k
)
(
x
)
g
(
k
)
(
x
)
)
′
=
∑
k
=
0
m
C
m
k
(
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
(
x
)
+
f
(
n
−
k
)
(
x
)
g
(
k
+
1
)
)
=
∑
k
=
0
m
C
m
k
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
(
x
)
+
∑
k
=
0
m
C
m
k
f
(
n
−
k
)
(
x
)
g
(
k
+
1
)
=
f
(
n
+
1
)
(
x
)
g
(
x
)
+
∑
k
=
1
m
C
m
k
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
(
x
)
+
∑
k
=
1
m
+
1
C
m
k
−
1
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
(
x
)
=
∑
k
=
0
m
C
m
k
(
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
(
x
)
+
f
(
n
−
k
)
(
x
)
g
(
k
+
1
)
)
=
f
(
n
+
1
)
(
x
)
g
(
x
)
+
∑
k
=
1
m
C
m
k
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
(
x
)
+
∑
k
=
1
m
C
m
k
−
1
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
+
C
m
m
f
(
x
)
g
(
m
+
1
)
(
x
)
=
f
(
n
+
1
)
(
x
)
g
(
x
)
+
(
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
(
x
)
)
(
∑
k
=
1
m
C
m
k
+
∑
k
=
1
m
C
m
k
−
1
)
+
f
(
x
)
g
(
m
+
1
)
(
x
)
(f(x)g(x))^{(m+1)}\\=((f(x)g(x))^{(m)})'\\ =\sum\limits_{k=0}^{m}\mathrm{C}_{m}^{k}(f^{(n-k)}(x)g^{(k)}(x))'\\ =\sum\limits_{k=0}^{m}\mathrm{C}_{m}^{k}(f^{(n-k+1)}(x)g^{(k)}(x)+f^{(n-k)}(x)g^{(k+1)})\\ =\sum\limits_{k=0}^{m}\mathrm{C}_{m}^{k}f^{(n-k+1)}(x)g^{(k)}(x)+\sum\limits_{k=0}^{m}\mathrm{C}_{m}^{k}f^{(n-k)}(x)g^{(k+1)}\\ =f^{(n+1)}(x)g(x)+\sum\limits_{k=1}^{m}\mathrm{C}_{m}^{k}f^{(n-k+1)}(x)g^{(k)}(x)+\sum\limits_{k=1}^{m+1}\mathrm{C}_{m}^{k-1}f^{(n-k+1)}(x)g^{(k)}(x)\\ =\sum\limits_{k=0}^{m}\mathrm{C}_{m}^{k}(f^{(n-k+1)}(x)g^{(k)}(x)+f^{(n-k)}(x)g^{(k+1)})\\ =f^{(n+1)}(x)g(x)+\sum\limits_{k=1}^{m}\mathrm{C}_{m}^{k}f^{(n-k+1)}(x)g^{(k)}(x)+\sum\limits_{k=1}^{m}\mathrm{C}_{m}^{k-1}f^{(n-k+1)}(x)g^{(k)}+\mathrm{C}_{m}^{m}f(x)g^{(m+1)}(x)\\ =f^{(n+1)}(x)g(x)+(f^{(n-k+1)}(x)g^{(k)}(x))(\sum\limits_{k=1}^{m}\mathrm{C}_{m}^{k}+\sum\limits_{k=1}^{m}\mathrm{C}_{m}^{k-1})+f(x)g^{(m+1)}(x)
(f(x)g(x))(m+1)=((f(x)g(x))(m))′=k=0∑mCmk(f(n−k)(x)g(k)(x))′=k=0∑mCmk(f(n−k+1)(x)g(k)(x)+f(n−k)(x)g(k+1))=k=0∑mCmkf(n−k+1)(x)g(k)(x)+k=0∑mCmkf(n−k)(x)g(k+1)=f(n+1)(x)g(x)+k=1∑mCmkf(n−k+1)(x)g(k)(x)+k=1∑m+1Cmk−1f(n−k+1)(x)g(k)(x)=k=0∑mCmk(f(n−k+1)(x)g(k)(x)+f(n−k)(x)g(k+1))=f(n+1)(x)g(x)+k=1∑mCmkf(n−k+1)(x)g(k)(x)+k=1∑mCmk−1f(n−k+1)(x)g(k)+Cmmf(x)g(m+1)(x)=f(n+1)(x)g(x)+(f(n−k+1)(x)g(k)(x))(k=1∑mCmk+k=1∑mCmk−1)+f(x)g(m+1)(x)
由公式
C
m
n
=
C
m
−
1
n
−
1
+
C
m
−
1
n
\mathrm{C}_{m}^{n}=\mathrm{C}_{m-1}^{n-1}+\mathrm{C}_{m-1}^{n}
Cmn=Cm−1n−1+Cm−1n可知
f
(
n
+
1
)
(
x
)
g
(
x
)
+
(
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
(
x
)
)
(
∑
k
=
1
m
C
m
k
+
∑
k
=
1
m
C
m
k
−
1
)
+
f
(
x
)
g
(
m
+
1
)
(
x
)
=
C
0
m
+
1
f
(
n
+
1
)
(
x
)
g
(
x
)
+
∑
k
=
1
m
C
m
+
1
k
(
f
(
n
−
k
+
1
)
(
x
)
g
(
k
)
(
x
)
)
+
C
m
+
1
m
+
1
f
(
x
)
g
(
m
+
1
)
(
x
)
=
∑
k
=
0
n
C
m
+
1
k
f
(
m
+
1
−
k
)
(
x
)
g
(
k
)
(
x
)
f^{(n+1)}(x)g(x)+(f^{(n-k+1)}(x)g^{(k)}(x))(\sum\limits_{k=1}^{m}\mathrm{C}_{m}^{k}+\sum\limits_{k=1}^{m}\mathrm{C}_{m}^{k-1})+f(x)g^{(m+1)}(x)\\ =\mathrm{C}_{0}^{m+1}f^{(n+1)}(x)g(x)+\sum\limits_{k=1}^{m}\mathrm{C}_{m+1}^{k}(f^{(n-k+1)}(x)g^{(k)}(x))+\mathrm{C}_{m+1}^{m+1}f(x)g^{(m+1)}(x)\\ =\sum\limits_{k=0}^{n}\mathrm{C}_{m+1}^{k}f^{(m+1-k)}(x)g^{(k)}(x)
f(n+1)(x)g(x)+(f(n−k+1)(x)g(k)(x))(k=1∑mCmk+k=1∑mCmk−1)+f(x)g(m+1)(x)=C0m+1f(n+1)(x)g(x)+k=1∑mCm+1k(f(n−k+1)(x)g(k)(x))+Cm+1m+1f(x)g(m+1)(x)=k=0∑nCm+1kf(m+1−k)(x)g(k)(x)
所以莱布尼茨公式对
n
=
m
+
1
n=m+1
n=m+1也成立
【注】
C
n
k
=
n
!
k
!
(
n
−
k
)
!
\mathrm{C}_{n}^{k}=\frac{n!}{k!(n-k)!}
Cnk=k!(n−k)!n!
【注】可以类比
(
a
+
b
)
n
=
∑
k
=
0
n
C
n
k
a
n
−
k
b
k
(a+b)^n=\sum\limits_{k=0}^{n}\mathrm{C}_{n}^{k}a^{n-k}b^k
(a+b)n=k=0∑nCnkan−kbk
【例4.5.5】求
y
=
(
3
x
2
−
2
)
sin
2
x
y=(3x^2-2)\sin 2x
y=(3x2−2)sin2x的
100
100
100阶导数。
【解】
y
(
100
)
=
∑
k
=
0
100
C
100
k
(
sin
2
x
)
(
100
−
k
)
(
3
x
2
−
2
)
(
k
)
y^{(100)}=\sum\limits_{k=0}^{100}\mathrm{C}_{100}^{k}(\sin 2x)^{(100-k)}(3x^2-2)^{(k)}
y(100)=k=0∑100C100k(sin2x)(100−k)(3x2−2)(k)
由于从
n
=
3
n=3
n=3开始，
(
3
x
2
−
2
)
(
n
)
=
0
(3x^2-2)^{(n)}=0
(3x2−2)(n)=0，所以
y
(
100
)
=
∑
k
=
0
2
C
100
k
(
sin
2
x
)
(
100
−
k
)
(
3
x
2
−
2
)
(
k
)
=
(
sin
2
x
)
(
100
)
(
3
x
2
−
2
)
+
100
(
sin
2
x
)
(
99
)
⋅
6
x
+
4950
(
sin
2
x
)
(
98
)
⋅
6
y^{(100)}=\sum\limits_{k=0}^{2}\mathrm{C}_{100}^{k}(\sin 2x)^{(100-k)}(3x^2-2)^{(k)}=(\sin 2x)^{(100)}(3x^2-2)+100(\sin 2x)^{(99)}\cdot 6x+4950(\sin 2x)^{(98)}\cdot 6
y(100)=k=0∑2C100k(sin2x)(100−k)(3x2−2)(k)=(sin2x)(100)(3x2−2)+100(sin2x)(99)⋅6x+4950(sin2x)(98)⋅6
(
sin
2
x
)
(
n
)
=
2
n
sin
(
2
x
+
n
π
2
)
(\sin 2x)^{(n)}=2^n\sin(2x + \frac{n\pi}{2})
(sin2x)(n)=2nsin(2x+2nπ)
所以
y
(
100
)
=
(
sin
2
x
)
(
100
)
(
3
x
2
−
2
)
+
100
(
sin
2
x
)
(
99
)
⋅
6
x
+
4950
(
sin
2
x
)
(
98
)
⋅
6
=
2
100
sin
(
2
x
+
100
π
2
)
(
3
x
2
−
2
)
+
100
⋅
2
99
sin
(
2
x
+
99
π
2
)
⋅
6
x
+
4950
⋅
2
98
sin
(
2
x
+
98
π
2
)
⋅
6
=
2
98
(
4
(
3
x
2
−
2
)
sin
2
x
−
1200
x
cos
2
x
+
29700
sin
2
x
)
=
2
98
(
4
(
3
x
2
−
2
)
sin
2
x
−
1200
x
cos
2
x
−
29700
sin
2
x
)
=
2
98
(
(
12
x
2
−
29708
)
sin
2
x
−
1200
x
cos
2
x
)
y^{(100)}=(\sin 2x)^{(100)}(3x^2-2)+100(\sin 2x)^{(99)}\cdot 6x+4950(\sin 2x)^{(98)}\cdot 6\\=2^{100}\sin(2x+\frac{100\pi}{2})(3x^2-2)+100\cdot 2^{99}\sin(2x+\frac{99\pi}{2})\cdot 6x+4950\cdot 2^{98}\sin(2x + \frac{98\pi}{2})\cdot 6\\ =2^{98}(4(3x^2-2)\sin 2x-1200x\cos 2x+29700\sin 2x)\\ =2^{98}(4(3x^2-2)\sin 2x-1200x\cos 2x-29700\sin 2x)\\ =2^{98}((12x^2-29708)\sin 2x-1200x\cos 2x)
y(100)=(sin2x)(100)(3x2−2)+100(sin2x)(99)⋅6x+4950(sin2x)(98)⋅6=2100sin(2x+2100π)(3x2−2)+100⋅299sin(2x+299π)⋅6x+4950⋅298sin(2x+298π)⋅6=298(4(3x2−2)sin2x−1200xcos2x+29700sin2x)=298(4(3x2−2)sin2x−1200xcos2x−29700sin2x)=298((12x2−29708)sin2x−1200xcos2x)
【注】
(
f
(
x
)
g
(
x
)
)
(
n
)
=
(
f
(
x
)
1
g
(
x
)
)
(
n
)
=
.
.
.
(\frac{f(x)}{g(x)})^{(n)}=(f(x)\frac{1}{g(x)})^{(n)}=...
(g(x)f(x))(n)=(f(x)g(x)1)(n)=...
复合函数，隐函数，参数表示的函数的高阶导数：
复合函数：
y
=
f
(
u
)
,
u
=
g
(
x
)
,
d
y
d
x
=
d
y
d
u
⋅
d
u
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y=f(u),u=g(x),\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
y=f(u),u=g(x),dxdy=dudy⋅dxdu
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y''(x)=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dx}(\frac{dy}{du}\cdot\frac{du}{dx})=\frac{d}{dx}(\frac{dy}{du})\cdot(\frac{du}{dx})+\frac{dy}{du}\cdot\frac{d}{dx}(\frac{du}{dx})=\frac{d}{du}(\frac{dy}{du})\cdot\frac{du}{dx}\cdot(\frac{du}{dx})+\frac{dy}{du}\cdot\frac{d}{dx}(\frac{du}{dx})=f''(u)(g'(x))^2+f'(u)g''(x)
y′′(x)=dxd(dxdy)=dxd(dudy⋅dxdu)=dxd(dudy)⋅(dxdu)+dudy⋅dxd(dxdu)=dud(dudy)⋅dxdu⋅(dxdu)+dudy⋅dxd(dxdu)=f′′(u)(g′(x))2+f′(u)g′′(x)
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y'''(x)=f'''(u)g'(x)\cdot (g'(x))^2+f''(u)2g'(x)g''(x)+f''(u)g'(x)g''(x)+f'(u)g'''(x)=f'''(u)(g'(x))^3+3f''(u)g'(x)g''(x)+f'(u)g'''(x)
y′′′(x)=f′′′(u)g′(x)⋅(g′(x))2+f′′(u)2g′(x)g′′(x)+f′′(u)g′(x)g′′(x)+f′(u)g′′′(x)=f′′′(u)(g′(x))3+3f′′(u)g′(x)g′′(x)+f′(u)g′′′(x)
原文地址：https://blog.csdn.net/qq_30204431/article/details/142750298
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