高阶导数与高阶微分

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由定义3.4知 : y(n)
x= x0
=
f
(n)
( x0
)
=
lim
∆x→0
f
(n−1) ( x0
+ ∆x) − ∆x
f
(n−1) ( x0 )
= lim f (n−1) (x) − f (n−1) (x0 ) .
x → x0
x − x0
y(n) = f (n) (x) = lim f (n−1) (x + ∆x) − f (n−1) (x)
2018/11/5
Edited by Lin Guojian
9
例 : 设y = ln(1+ x),求y(n).
解 : y′ = [ln(1+ x)]′ = 1 = (1+ x)−1 1+ x
y′′ = [ln(1+ x)]′′ = [(1+ x)−1]′ = (−1)(1+ x)−2
y′′′ = [ln(1+ x)]′′′ = [(−1)(1+ x)−2 ]′ = (−1)(−2)(1+ x)−3 设y(k) = [ln(1+ x)](k) = (−1)(−2)(−k +1)(1+ x)−k = (−1)k−1(k −1)!(1+ x)−k
解 : y(n) = a0n!
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Edited by Lin Guojian
5
例 : 设y = ex , 求y(n)与y(n) (0).
解 : y′ = (ex )′ = ex , y′′ = (ex )′ = ex ,, y(n) = ex.
y(n) (0) = y(n) = ex = e0 = 1.

:
lim
∆x→0
f ′(x0
+ ∆x) − ∆x
f ′(x0 )
存在, 则称y
=
f
(x)在
点 x0 处二阶可导, 且称y′ = f ′(x)在点 x0 处的导数为函数f (x)在点 x0 的二阶导数.
用f
′′(x0 ) 或 y′′
x= x0
d2y , 或 dx2
x= x0
d2 f , 或 dx2
x=0
x=0
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Edited by Lin Guojian
6
例 : 设y = sin x, 求y(n).
解 : y′ = (sin x)′ = cos x = sin(x + π ),
2
y′′ = (cos x)′ = − sin x = sin(x + π ) = sin(x + 2 ⋅ π ).
y(k) = k(k −1)(k − 2)3⋅ 2 ⋅1 = k!
y(k+1) = 0, y(k+2) = 0,, y(k+i) = 0(i ≥ 1).
( ) 故y(n) =
xk
(n)
k(k −1)(k − n +1)xk−n , =
n≤k
0
n>k
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Edited by Lin Guojian
22
2
由数学归纳法知 : (cos x)(n) = cos[x + n ⋅ π ], (n = 1,2,3,).
2
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Edited by Lin Guojian
8
例 : 设y = x−1, 求y(n).
解 : y′ = [x−1]′ = −x−2. y′′ = [(−x−2 ]′ = (−1)(−2) ⋅ x−3. y′′′ = [(−1)(−2) ⋅ x−3]′ = (−1)(−2)(−3) ⋅ x−4. 设y(k) = (−1)(−2)(−k ) ⋅ x−k−1 = (−1)k ⋅ k!⋅x−(k+1).
y′′ = (ex + xex )′ = ex + ex + xex = 2ex + xex.
例: 设 y = arctan x, 求y′′, y′′′.

:
y′
=
1 1+ x2
y′′
=
1
1
+
x
2

=
− 2x (1+ x2 )2
.
y′′′
=
− 2x
(1
+
x
2
)
2

=
2(3x2 −1) (1+ x2 )3
3.5、高阶导数与高阶微分
设y = f (x)在(a,b)内可导,则它的导函数y′ = f ′(x)和微分函数dy = df (x) = f ′(x)dx仍然是 (a,b)上的函数,因此可以继续讨论它们的可导性与可微性, 这就产生了高阶与高阶微分.
一、高阶导数
定义3.4 : 如果y′ =
f
′(
x)在x0点处可导,
2018/11/5
Edited by Lin Guojian
12
例 : 设y = x3e−x ,求y(30).
30
∑ 解 : [x3e− x ](30) =
C3k0 f (k)(x)g(30−k)(x)
k =0
= C300 f (0)(x)g(30)(x) + C310 f ′(x)g(29)(x) + C320 f ′′(x)g(28)(x) + C330 f ′′′(x)g(27)(x)
2 设y(k) = (sin x)(k) = sin(x + k ⋅ π ).
2
则y(k+1) = (sin x)(k+1) = [sin(x + k ⋅ π )]′ = cos(x + k ⋅ π )
2
2
= sin(x + k ⋅ π + π ) = sin[x + (k +1) ⋅ π ].
22
+ C320 f ′′(x)g(28)(x) + C330 f ′′′(x)g(27)(x)
= C300 x3e−x + C3103x2(−1)e−x + C320 6xe−x + C330 6(−1)e−x = x3e−x − 90x2e−x + 2610xe−x − 24360e−x .
2018/11/5
则y(k+1) = [ln(1+ x)](k+1) = [(−1)k−1(k −1)!(1+ x)−k ]′
= (−1)k−1(k −1)!(−k)(1+ x)−k−1 = (−1)k (k)!(1+ x)−(k+1).
由数学归纳法知 : y(n) = (−1)n−1(n −1)!(1+ x)−n , (n = 1,2,3,).
Edited by Lin Guojian
13
例 : 求y = arctan x在x = 0点的n阶导数.
解 : y′ = 1 , y′′ = − 2x , y′′′ = − 2 + 6x2 .
1 + x2
(1 + x2 )2
(1 + x2 )3
由y′ = 1 ⇒ (1 + x2)y′ = 1. 1 + x2
y′′
=
(−
sin
x)′
=

cos
x
=
cos( x
2 +
π
)
=
cos(
x
+
2

π
)
2
设y(k) = (cos x)(k) = cos(x + k ⋅ π )
2
则y(k+1) = (cos x)(k+1) = [cos(x + k ⋅ π )]′ = − sin(x + k ⋅ π )
2
2
= cos(x + k ⋅ π + π ) = cos[x + (k +1) ⋅ π ].
2018/11/5
Edited by Lin Guojian
10
例 : 设y = ln( x2 + 3x − 4), 求y(n).
解 : y = ln( x −1) + ln( x + 4) ⇒ y′ = 1 + 1 . x −1 x + 4
y′′
=
1 x −1
+
x
1 +
′ 4
=
1

x −1
[ ] 在(1 + x2)y′ = 1方程两边对x求n − 1阶导数,即(1 + x2)y′ (n−1) = 1(n−1).
[ ] ∑ 那么 : (1 +
x2 )y′ (n−1)
=
n −1
Cnk−1(1 +
x2 )(k )( y′)(n−1−k )
k =0
= Cn0−1(1 + x2 )(0)( y′)(n−1) + Cn1−1(1 + x2 )(1)( y′)(n−2) + Cn2−1(1 + x2 )(2)( y′)(n−3)
.
2018/11/5
Edited by Lin Guojian
3
例 : 求y = xk (k为正整数)的n阶导数y(n).
解 : y′ = (xk )′ = kxk−1 y′′ = (kxk−1)′ = k (k −1)xk−2
y′′′ = [k(k −1)xk−2 ]′ = k(k −1)(k − 2)xk−3 y(k−1) = k(k −1)(k
n! , k!(n − k)!
f (0)(x) = f(x).
注 : (a) : 规定 : 0! = 1,Cn0 = 1. (b) : 如果f(x) = xk ,则当n > k时, f (n)(x) = 0, 此时应用莱布尼茨公式计算[ f(x) ⋅ g(x)](n)时可能有很多项为零.
2
由数学归纳法知 : (sin x)(n) = sin[x + n ⋅ π ], (n = 1,2,3,).
2
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Edited by Lin Guojian
7
例 : 设y = cos x, 求y(n).
解 : y′ = (cos x)′ = − sin x = cos(x + π )
= (−1)n−1(n −1)!(x −1)−n + (−1)n−1(n −1)!(x + 4)−n.
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Edited by Lin Guojian
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求函数的高阶导数常用以下两个公式 :
(1) : [ f(x) + g(x)](n) = f (n)(x) + g(n)(x);
n
∑ (2) : (莱布尼茨公式) : [ f(x) ⋅ g(x)](n) = Cnk f (k)(x)g(n−k)(x), k =0
= (1 + x2)y(n) + (n − 1)2xy(n−1) + (n − 1)(n − 2)y(n−2) = 0.
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Edited by Lin Guojian
14
令x = 0可得 : y(n)(0) + (n − 1)(n − 2)y(n−2)(0) = 0 ⇒ y(n)(0) = −(n − 1)(n − 2)y(n−2)(0).
+
C340
f (4)(x)g(26)(x) +
+
C 30 30
f ( (30) x)g(0)(x).
由于 : (x3 )(k) = 0, k > 3,且(e−x )(n) = (−1)n e−x .
因此 : [x3e−x ](30) = C300 f (0)(x)g(30)(x) + C310 f ′(x)g(29)(x)
+
x
1 +
′ 4
=
(−1)( x
− 1) − 2
+
(−1)( x
+
4) − 2 .
y′′′ = (−1)(−2)(x −1)−3 + (−1)(−2)(x + 4)−3.
y(n) = (−1)(−2)[−(n −1)]( x −1)−n + (−1)(−2)[−(n −1)]( x + 4)−n
则y(k+1) = [(−1)k ⋅ k!⋅x−(k+1) ]′ = (−1)k ⋅ k!⋅[−(k +1)]⋅ x−(k+2) = (−1)k+1 ⋅ (k +1)!⋅x−(k+2).
由数学归纳法知 : y(n) = (x−1)(n) = (−1)n n!x−(n+1) , (n = 1,2,).
表示f (x)在x0的二阶导数. x = x0
即:
f
′′(x0 )
=
lim
∆x→0
f ′(x0
+ ∆x) − ∆x
f
′(x0 )
2018/11/5
Edited by Lin Guojian
1
类似地可以定义三阶导数f ′′′(x0 ),四阶导数f (4) (x0 )及n阶导数f (n) (x0 ).
由于y′(0) = 1, y′′(0) = 0, 因此,当n为偶数,即n = 2k,k = 1,2, . 则y(4)(0) = −(4 − 1() 4 − 2)y(2)(0) = 0,
y(6)(0) = −(6 − 1() 6 − 2)y(4)(0) = 0, , y(2k)(0) = −(2k − 1() 2k − 2)y(2k −2)(0) = 0.
∆x→0
∆x
即 : f (n) (x) = [ f (n−1) (x)]′ = df (n−1) (x) , n = 1,2,3,, dx
通常称二阶及二阶以上的导数为高阶导数.
2018/11/5
Edited by Lin Guojian
2
例: 设y = f (x) = xex ,求y′′.
解 : y′ = (xex )′ = ex + xex
4
例 : 设y = 2x4 + 4x3 + 7x2 + 8x +1, 求y(4). 解 : y(4) = (2x4 + 4x3 + 7x2 + 8x +1)(4) = 2 ⋅ 4!= 48.
例: 求n次多项式 y = a0 xn + a1xn−1 + + an−1x + an , (a0 ≠ 0)的n阶导数.
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