18 Higher Derivative of the Product of Two Functions
18.1 Leibniz Rule about the Higher Order Differentiation
Theorem 18.1.1 (Leibniz)
When functions f( )x and g( )x are n times differentiable, the following expression holds.
f( )x g( )x ( )n = Σr=0
n
n
rf ( )n-r ( )x g( )r ( )x (1.1)
Proof Theorem 16.1.2 (2.1) in 16.1.2 was as follows.
an
x
a1
x
f < >0 g( )0 dxn = Σr=0
m -1
-n
rf < >n+ r g( )r
- Σr=0
n -1
Σs=0
m -1
-n + r
sf < >n-r+ s
an-r g( )s
an-ran
x
an- r+1
x
dxr
+ ( )-1 m Σr=1
n -1
Σs=0
r-1
Σt=s
r-1
Ct s C m+n-1-r+ t m-1 f m+ n-r+san-r
g( )m+ s an-r an
x
an- r+1
x
dxr
+ ( )n ,m( )-1 m
Σk=0
n -1
m+kCn -1 k
an
x
a1
x
f< >m+ k g( )m+ k dxn
Should be noted here is the next two.
i When n =1 ∑∑∑ of the 3rd line does not exist, when n =0 ∑∑ of the 2nd line does not exist also.
ii When the binomial coefficient of the 4th line is generalized, the upper limit n-1 of ∑ can be replaced
by .
Since 1>0 -n at the time n =0,1,2, , if the index n of the integration operator is substituted for
-n in consideration of these, it becomes as follows.
an
x
a1
x
f < >0 g( )0 dx-n = Σr=0
m -1
n
rf < >-n+ r g( )r
+ ( )-n ,m( )-1 m
Σk=0
m+k1
-n -1
k an
x
a1
x
f< >m+ k g( )m+ k dx-n
Since m may be arbitrary integer, when m=n +1 , it is as follows.
an
x
a1
x
f < >0 g( )0 dx-n = Σr=0
n
n
rf < >-n+ r g( )r
+ ( )-n ,n +1( )-1 n+1
Σk=0
n +1+k1
-n -1
k an
x
a1
x
f< >n+1+ k g( )n+1+ k dx-n
However, since ( )-n ,n +1 = for n =0,1,2, , the 2nd line disappears. That is,
an
x
a1
x
f g dx-n = Σr=0
n
n
rf< >-n+ r g( )r n =0,1,2,
Then, replacing the integration operators dx -n ,< >-n +r with the differentiation operators ( )n ,( )n -rrespectivly, we obtain the desired expression.
- 1 -
18.2 Higher Derivative of x̂ a f (x)
Formula 18.2.0
When ( )z denotes the gamma function and f( )x is n times differentiable continuous function,
the following expressions hold for a natural number n.
(1)
x f( )x( )n
= Σr=0
n
n
r ( )1+-r( )1+
x- rf( )n- r ( )x (0.1)
Where, if = -1,-2,-3,, it shall read as follows.
1+-r
( )1+ ( )-1 -r
( )-( )-+r
(2) Especially, when =m =0,1,2,
xm f( )x( )n
= Σr=0
m
n
r ( )1+m-r( )1+m
xm- rf( )n- r ( )x (0.1')
(3) When -1,-2,-3, & -n -1,-2,-3,
x f( )x( )n
= Σr=0
n
n
r ( )1+-n+r( )1+
x-n+ rf( )r ( )x (0.2)
Proof
When g( )x = x in Theorem 18.1.1 , since
x ( )r = ( )1+-r
( )1+ x- r
we obtain the following expression immediately.
x f( )x( )n
= Σr=0
n
n
r ( )1+-r( )1+
x- r f( )n- r ( )x (0.1)
Especially, when =m =0,1,2, , (0.1) is as follows.
x m f( )x( )n
=Σr=0
n
n
r ( )1+m-r( )1+m
x m- r f( )n- r ( )x ( )1+m-r( )1+m
=0 for m<rn
=Σr=0
m
n
r ( )1+m-r( )1+m
x m- r f( )n- r ( )x n
r=0 for n <rm
We adopt the convenient latter for mathematical software.
When = -1,-2,-3, , from 1.1.5 ( Properties of the Gamma Function ) (5.5),
( )-z-n( )-z
= ( )-1 -n
( )1+z( )1+z+n
(n is a non-negative integer )
Then substituting -z = 1+ , n= r for this, we obtain the proviso.
Last, replacing r with n -r in (0.1), we obtain (0.2).
Below, substituting various functions f for Formula 18.2.0 , we obtain various formulas.
Although there are (1) and (2) in Formula 18.2.0, since (2) is almost meaningless in the case of higher
differentiation, we adopt (1) in principle.
- 2 -
18.2.1 Higher Derivative of ( )ax+b p( )cx+d q
Formula 18.2.1
The following expressions hold for p >0 and n =1,2,3, .
( )ax+b p( )cx+d q ( )n
= Σr=0
n
n
r ( )1/c r
( )1/a -n+ r
( )1+p -n+r ( )1+q -r( )1+p ( )1+q
( )cx+d r-q
( )ax+b p-n+r
(1.1)
Especially, when m =0,1,2,
( )ax+b p( )cx+d m ( )n
= Σr=0
m
n
r ( )1/c r
( )1/a -n+ r
( )1+p -n+r ( )1+m-r( )1+p ( )1+m
( )cx+d r-m
( )ax+b p-n+ r
(1.1')
Proof
Let f( )x =( )ax+b p , g( )x =( )cx+d q, then
f( )n- r = ( )ax+b p ( )n-r = a
1 -n+r
( )1+p -n +r( )1+p
( )ax+b p-n+r
g( )r = ( )cx+d q ( )r= c
1 -r
( )1+q-r( )1+q
( )cx+d q-r q 1, 2, 3,
Substituting these for Theorem 18.1.1 , we obtain (1.1) .
And especially, when q = m = 0,1,2, , from (1.1)
( )ax+b p( )cx+d m ( )n
= Σr=0
n
n
r ( )1/c r
( )1/a -n+ r
( )1+p -n+r ( )1+m-r( )1+p ( )1+m
( )cx+d r-m
( )ax+b p-n+r
= Σr=0
m
n
r ( )1/c r
( )1/a -n+ r
( )1+p -n+r ( )1+m-r( )1+p ( )1+m
( )cx+d r-m
( )ax+b p-n+r
We adopt the latter expression as (1.1').
Example1 The 2nd order derivative of x-2 3 3x+4
Substituting a =1, b=-2, p =1/2 , c=3, d =4, q=1/3 , n =2 for (1.1) ,
x-2 3 3x+4( )2
=Σr=0
2
2
r3r
( )r-1/2 ( )4/3-r( )3/2 ( )4/3
( )x-2r- 2
3
( )3x+4 31
- r
= - x-2 3 3x+4 4( )x-2 2
1-
( )x-2 ( )3x+41
+( )3x+4 2
2
Example1' The 3rd order derivative of x-2 ( )3x+4 2
Substituting a =1, b=-2, p =1/2 , c=3, d =4, m=2 , n=3 for (1.1') ,
- 3 -
x-2 ( )3x+4 2 ( )3 = Σ
r=0
2
3
r3r
( )-3/2+r ( )3-r( )3/2 ( )3
( )3x+4 r-2
( )x-2- 2
5+ r
= x -2( )3x +4 2 8( )x -2 3
31
- 2( )3x +4 ( )x -2 2
32
+ ( )3x +4 2( )x -2
33
Example2 The 3rd order derivative of x-2 / ( )3x+4
When q = -1,-2,-3, , (1.1) can be read as follows.
( )ax+b p( )cx+d q ( )n
= Σr=0
n
n
r ( )-1/c r
( )1/a -n+ r
( )1+p -n +r ( )-q( )1+p ( )-q+r
( )cx+d r-q
( )ax+b p-n+ r
Substituting a=1, b =-2, p=1/2 , c=3, d =4, q =-1 , n=3 for this,
3x+4x-2
( )3
=Σr=0
3
3
r( )-3 r
( )-3/2+r ( )1( )3/2 ( )1+r
( )3x+4 r+1
( )x-2- 2
5+ r
= 3x +4x -2
8( )x -2 3
3+
4( )x -2 2( )3x +4
9+
( )x -2 ( )3x +4 2
27-
( )3x +4 3
162
18.2.2 Higher Derivative of x log x
Formula 18.2.2
x log x( )n
= -Σr=0
n -1
( )-1 n-r n
r ( )1+-r( )n -r ( )1+
x -n + ( )1+-n( )1+
x -n log x(2.1)
Especially, when m = 0,1,2,
x m log x( )n
= -Σr=0
n -1
( )-1 n-r n
r ( )1+m-r( )n -r ( )1+m
x m-n + ( )1+m-n( )1+m
x m-n log x(2.1')
Where, there shall be no 2nd term of the right side at the time of m< n .
Proof
Let f( )x = log x . Then
( )log x ( )n-r = -( )-1 n-r( )n -r x-n+r r = 0,1,, n -1 = log x r = n
Substituting these for (0.1) in Theorem 18.2.0 , we obtain (2.1).
When m= 0,1,2, , applying (0.1') , we obtain the following.
x m log x( )n
= -Σr=0
n -1
( )-1 n-r n
r ( )1+m-r( )n -r ( )1+m
x m-n + ( )1+m-n( )1+m
x m-n log x
+Σr=n +1
m
n
r ( )1+m-r( )1+m
x m-n( )log x ( )n-r
Where, since r does not reach n at the time of m< n , the 2nd term does not exist.
- 4 -
Example1 The 3rd order derivaive of x log x
Substituting =1/2 , n=3 for (2.1) ,
x log x( )3
= -Σr=0
2
( )-1 3-r 3
r ( )3/2-r( )2-r ( )3/2
x-
25
+ ( )-3/2( )3/2
x-
25
log x
= - - 3
02!
( )3/2( )3/2
+ 3
11!
( )1/2( )3/2
- 3
20!
( )-1/2( )3/2
x-
25
+ ( )-3/2( )3/2
x-
25
log x
= 2-23
-43
x-
25
+
( )-3/2( )3/2
x-
25
log x = -
41
+83
log x x- 2
5
Example1' The 2nd order derivaive of x3log x
Substituting m=3 , n=2 for (2.1') ,
x3log x( )2
= -Σr=0
1
( )-1 2-r 2
r ( )4-r( )2-r ( )4
x1 + ( )2( )4
x1log x
= - ( )-1 2 2
0 3!1!3!
+ ( )-1 1 2
1 2!0!3!
x1 + 1!3!
x1 log x
= ( )5 + 6 log x x1
Example1" The 3rd order derivaive of x2log x
Substituting m=2 , n=3 for (2.1') ,
x2log x( )3
= -Σr=0
2
3
r( )-1 3-r
( )3-r( )3-r ( )3
x-1
= - 3
0( )-1 32! +
3
1( )-1 22! +
3
2( )-1 12! x-1
= 2 x-1
Example2 The 3rd order derivaive of log x /x
When = -1,-2,-3, , (2.1) can be read as follows.
x log x( )n
= -( )-1 nΣr=0
n -1
n
r ( )-( )n -r ( )-+r
x -n + ( )-1 n
( )-( )-+n
x -n log x
Substituting =-1 , n =3 for this ,
xlog x ( )3
= -( )-1 -3Σr=0
2
3
r ( )1( )3-r ( )1+r
x-4 + ( )-1 -3
( )1( )4
x-4log x
= x -4 3
0( )3 ( )1 +
3
1( )2 ( )2 +
3
2( )1 ( )3 - 6 log x
= x4
1( )11 - 6log x
- 5 -
18.2.3 Higher Derivatives of x sinx , x cosx
Formula 18.2.3
x sin x{ }n
=Σr=0
n
n
r ( )1+-r( )1+
x-r sin x+2
( )n-r (3.1s)
x cosx{ }n
=Σr=0
n
n
r ( )1+-r( )1+
x-r cos x+2
( )n -r (3.1c)
Especially, when m = 0,1,2,
xm sin x{ }n
=Σr=0
m
n
r ( )1+m-r( )1+m
xm-r sin x+2
( )n -r (3.1's)
xm cosx{ }n
=Σr=0
m
n
r ( )1+m-r( )1+m
xm-r cos x+2
( )n -r (3.1'c)
Example1 The 2nd order derivative of 3
x sinx
Substituting =1/3 , n =2 for (3.1s) ,
3x sin x
( )2= Σ
r=0
2
2
r 4/3-r( )4/3
x 31
-rsin x+
2( )2-r
= 2
0 4/3 4/3
x 31
sin x + + 2
1 1/3 4/3
x-
32
sin x +2 +
2
2 -2/3 4/3
x-
35
sin x
= -x 31
sin x +32
x- 3
2
cosx - 92
x- 3
5
sin x
Example1' The 3rd order derivative of x2sinx
Substituting m=2 , n=3 for (3.1's) ,
x2 sin x{ }3
=Σr=0
2
3
r 3-r( )3
x2-r sin x+2
( )3-r
= 3
0 3( )3
x 2 sin x +2
3 +
3
1 2( )3
x 1 sin x +2
2
+ 3
2 1( )3
x 0 sin x +2
= -x2 cosx - 6 x sin x + 6 cosx
18.2.4 Higher Derivatives of x sinhx , x coshx
Formula 18.2.4
x sinh x( )n
= Σr=0
n
n
r ( )1+-r( )1+
x- r
2ex -( )-1 -( )n-r e-x
(4.1s)
x cosh x( )n
= Σr=0
n
n
r ( )1+-r( )1+
x- r
2ex +( )-1 -( )n-r e-x
(4.1c)
- 6 -
Especially, when m = 0,1,2,
xm sinh x( )n
= Σr=0
m
n
r ( )1+m-r( )1+m
xm- r
2ex -( )-1 -( )n-r e-x
(4.1's)
xm cosh x( )n
= Σr=0
m
n
r ( )1+m-r( )1+m
xm- r
2ex +( )-1 -( )n-r e-x
(4.1'c)
Example1 The 2nd order derivaive of 3
x sinhx
Substituting =1/3 , n =2 for (4.1s) ,
3x sinh x
( )2= Σ
r=0
2
2
r 4/3-r( )4/3
x 31
-r
2ex -( )-1 -( )2-r e-x
= 2
0 4/3( )4/3
x 31
sinh x + 2
1 1/3( )4/3
x-
32
coshx + 2
2 -2/3( )4/3
x-
35
sinh x
= x 31
sinh x + 32
x- 3
2
cosh x - 92
x- 3
5
sinh x
Example1' The 3rd order derivative of x2sinhx
Substituting m=2 , n=3 for (4.1's) ,
x2 sinh x{ }3
=Σr=0
2
3
r 3-r( )3
x2-r
2ex -( )-1 -( )3-r e-x
= 3
0 3( )3
x 2 cosh x + 3
1 2( )3
x 1 sinh x + 3
2 1( )3
x 0 cosh x
= x2 cosh x + 6 x sinh x + 6 cosh x
- 7 -
18.3 Higher Derivative of log x f (x)
18.3.1 Higher Derivative of ( )log x 2
Formula 18.3.1
log 2x( )n
= xn
( )-1 n-1
2( )n log x -Σr=1
n -1
n
r( )n -r ( )r (1.1)
Proof
Let f( )x = g( )x = log x . Then
( )log x ( )n- r = ( )-1 n- r-1
x n- r
( )n -r , ( )log x ( )r = ( )-1 r-1
x r
( )rn =1,2,
Substituting these for Theorem 18.1.1 ,
log 2x( )n
= Σr=0
n
n
r( )log x ( )n- r ( )log x ( )r
= n
0( )log x ( )n ( )log x ( )0 + Σ
r=1
n -1
n
r( )log x ( )n- r ( )log x ( )r
+ n
n( )log x ( )n- n ( )log x ( )n
= ( )-1 n-1
xn
2( )n log x +
xn
( )-1 n
Σr=1
n -1
n
r( )n -r ( )r
= xn
( )-1 n-1
2( )n log x -Σr=1
n -1
n
r( )n -r ( )r
Example The 3rd order derivative of ( )log x 2
log 2x( )3
= x3
( )-1 3-1
2( )3 log x -Σr=1
2
3
r( )3-r ( )r
= x3
1 22log x -
3
1( )2 ( )1 -
3
2( )1 ( )2
= x3
1( )4log x -6
18.3.2 Higher Derivatives of log xsinx , log xcosx
Formula 18.3.2
( )log xsin x( )n
= log xsin x+2
n
+Σr=1
n
( )-1 r-1 n
r xr
( )rsin x+
2( )n -r
(2.0s)
- 8 -
( )log xcos x( )n
= log xcos x+2
n
+Σr=1
n
( )-1 r-1 n
r xr
( )rcos x+
2( )n -r
(2.0c)
Example The 3rd order derivative of log xsin x
( )log xsin x( )3 = log xsin x+
23
+Σr=1
3
( )-1 r-1 3
r xr
( )rsin x+
2( )3-r
= -log xcosx + 3
1 x1
( )1sin x+
22
- 3
2 x2
( )2sin x+
21
+ 3
3 x3
( )3sin x+
20
= -log xcosx - x1
3sin x -
x2
3cosx +
x3
2sin x
18.3.3 Higher Derivatives of log xsinhx , log xcoshx
Formula 18.3.3
log xsinh x( )n
= log x 2ex -( )-1 -ne-x
+Σr=1
n
( )-1 r-1 n
r xr
( )r2
ex -( )-1 r-ne-x
(3.0s)
log xcosh x( )n = log x 2
ex +( )-1 -ne-x
+Σr=1
n
( )-1 r-1 n
r xr
( )r2
ex +( )-1 r-ne-x
(3.0c)
Example The 4th order derivative of log xcosh x
( )log xcosh x( )4
= log x 2e x+( )-1 -4e -x
+Σr=1
4
( )-1 r-1 4
r x r
( )r2
e x+( )-1 r-4e -x
= log xcosh x + 4
1 x1
( )1sinh x -
4
2 x2
( )2cosh x
+ 4
3 x3
( )3sinh x -
4
4 x4
( )4cosh x
= log xcosx +x1
4sinh x -
x2
6cosh x +
x3
8sinh x -
x4
6cosh x
- 9 -
18.4 Higher Derivative of e x̂ f (x)
18.4.1 Higher Derivative of ex x
Formula 18.4.1
ex x( )n
= exΣr=0
n
n
r ( )1+-r( )1+
x- r for -1,-2,-3, (1.1)
= exΣr=0
n( )-1 r
n
r ( )-( )-+r
x- r for = -1,-2,-3, (1.2)
Especially, when m = 0,1,2,
ex xm ( )n = exΣ
r=0
m
n
r ( )1+m-r( )1+m
xm- r(1.1')
Proof
Substite f( )x = e xfor Theorem 18.2.0 . Then since e x ( )n-r
= e x, we obtain the desired expression
immediately.
Example1 The 2nd order derivative of e x x
ex x( )2
= exΣr=0
2
2
r 3/2-r( )3/2
x 21
- r
= e x 2
0 3/2( )3/2
x 21
+ 2
1 1/2( )3/2
x-
21
+
2
2 -1/2( )3/2
x-
23
= ex x 21
+ x
- 21
- 4
1x
- 23
= ex x 1+x1
-4x2
1
Example2 The 2nd order derivative of e x/x
xex ( )2
= exΣr=0
2( )-1 r
2
r ( )1( )1+r
x1- r
= ex 2
0 ( )1( )1
x-1 - 2
1 ( )1( )2
x-2 + 2
2 ( )1( )3
x-3
= xex
1 - x2
+ x2
2
18.4.2 Higher Derivative of ex log x
Formula 18.4.2
ex log x( )n
= ex log x + exΣr=1
n
( )-1 r-1 n
r xr
( )r(2.1)
Proof
Let f( )x = e x , g( )x = log x . Then
- 10 -
( )log x ( )r = ( )-1 r-1
xr
( )rr = 1,2,3,
Substituting this for Theorem 18.1.1 ,
ex log x( )n
= Σr=0
n
n
rex ( )log x ( )r =
n
0ex ( )log x ( )0 +Σ
r=1
n
ex ( )log x ( )r
= ex log x + exΣr=1
n
( )-1 r-1 n
r xr
( )r
Example The 4th order derivative of e xlog x
ex log x( )4
= ex log x + exΣr=1
4
( )-1 r-1 4
r xr
( )r
= ex log x + ex 4
1 x1
( )1-
4
2 x2
( )2+
4
3 x3
( )3-
4
4 x4
( )4
= ex log x + ex x1
4 -
x2
6 +
x3
8 -
x4
6
18.4.3 Higher Derivatives of ex sinx , ex cosx
Formula 18.4.3
exsin x( )n
= sin 4 -n
exsin x+4
n(3.0s)
excosx( )n
= sin 4 -n
excos x+4
n(3.0c)
Proof
"共立 数学公式" p187 was posted as it was.
Example
exsin x( )2
= sin 4 -2
exsin x+4
2 = 2ex cosx
excosx( )3
= sin 4 -3
excos x+4
3 = -2ex( )sin x+ cosx
Higher Derivatives of e xsin x, e xcos x end now. There is no necessity for Theorem 18.1.1.
However, daring use Theorem 18.1.1, we obtain an interesting result.
Trigonometric Polynomial
Formula 18.4.3'
Σr=0
n
n
rsin x+
2r
= sin 4 -n
sin x+4
n(3.1s)
- 11 -
Σr=0
n
n
rcos x+
2r
= sin 4 -n
cos x+4
n(3.1c)
Especially, when x=0
Σr=0
n
n
rsin 2
r = sin 4
-n
sin 4n
(3.1's)
Σr=0
n
n
rcos 2
r = sin 4
-n
cos 4n
(3.1'c)
Proof
Substituting f( )x = ex , g( )x = sin x, cosx for Theorem 18.1.1 ,
ex sin x( )n
= exΣr=0
n
n
rsin x+
2r
ex cos x( )n
= exΣr=0
n
n
rcos x+
2r
And comparing these with Formula 18.4.3 , we obtain the desired expressions.
When n =5 , if both sides of (3.1s) are illustrated, it is as follows. Both overlap exactly and blue (left) can
not be seen.
Alternative Binomial Polynomial
Removing sin2r
, cos2r
from (3.1's), (3.1'c), we obtain the following interesting polynomial.
Formula 18.4.3"
When denotes the floor function, the following expressions hold.
Σr=0
( )n -1 /2( )-1 r
n
2r+1 = 2 2
n
sin 4n
(3.2s)
Σr=0
n /2( )-1 r
n
2r = 2 2
n
cos 4n
(3.2c)
- 12 -
Proof Since the odd-numbered terms of the left side in (3.1's) are all 0,
Σr=0
n
n
rsin 2
r =
n
0sin 2
0+
n
1sin 2
1+
n
2sin 2
2++
n
nsin 2
n
= n
1sin
21
+ n
3sin
23
+ n
5sin
25
+ n
2n-1
sin2
2n-1
= n
1 -
n
3 +
n
5 -
n
2n-1
= Σr=0
( )n -1 /2( )-1 r
n
2r+1
Also, since sin /4 -n = 2n/2
in the right side in (3.1's) ,
Σr=0
( )n -1 /2( )-1 r
n
2r+1 = 2 2
n
sin 4n
(3.2s)
Next, since the even-numbered terms of the left side in (3.1'c) are all 0,
Σr=0
n
n
rcos 2
r =
n
0cos 2
0+
n
1cos 2
1+
n
2cos 2
2++
n
ncos 2
n
= n
0cos
20
+ n
2cos
22
+ n
4cos
24
+ n
n/2cos
2n /2
= n
0 -
n
2 +
n
4 -
n
n /2 = Σ
r=0
n /2( )-1 r
n
2r
Also, since sin /4 -n = 2n/2
in the right side in (3.1'c) ,
Σr=0
n /2( )-1 r
n
2r = 2 2
n
cos 4n
(3.2c)
In addition, this formula is known. (See "岩波 数学公式Ⅱ" p11)
Note
When n =4k -3 , k =1,2,3,
Σr=0
( )n -1 /2( )-1 r
n
2r+1 = Σ
r=0
n /2( )-1 r
n
2r
Example
5
1 -
5
3 +
5
5 = 5 - 10 + 1 = 2 2
5
sin 45
= -4
5
0 -
5
2 +
5
4 = 1 - 10 + 5 = 2 2
5
cos 45
= -4
18.4.4 Higher Derivatives of ex sinhx , ex coshx
Formula 18.4.4
exsinh x( )n
= exΣr=0
n
n
r 2ex - ( )-1 -re-x
(4.0s)
- 13 -
excosh x( )n
= exΣr=0
n
n
r 2ex + ( )-1 -re-x
(4.0c)
Example
ex sinh x( )0
= exΣr=0
0
0
r 2ex - ( )-1 -re-x
= exsinh x
ex cosh x( )3
= exΣr=0
3
3
r 2ex + ( )-1 -re-x
= ex 3
0sinh x+
3
1cosh x+
3
2sinh x+
3
3cosh x
= 4ex( )sinh x+ cosh x
Note
The following formula is known for a natural number n .
ex sinh x( )n
= ex cosh x( )n
= 2n-1ex( )sinh x+ cosh xHowever, this formula does not hold for n =0 . That is, in this formula, the natural number n is inextensible
to the real number p . So, this is insufficient as a general formula.
- 14 -
18.5 Higher Derivative of f (x) / e x̂
18.5.1 Higher Derivative of e-x x
Formula 18.5.1
e -x x ( )n = e -x
Σr=0
n
( )-1 -( )n-r n
r ( )1+-r( )1+
x - r for -1,-2,-3, (1.1)
= ( )-1 -ne -xΣr=0
n
n
r ( )-( )-+r
x - r for = -1,-2,-3, (1.2)
Especially, when m = 0,1,2,
e-x xm ( )n = e-x
Σr=0
m
( )-1 -( )n-r n
r ( )1+m-r( )1+m
xm- r(1.1')
Proof
Substite f( )x = e -xfor Theorem 18.2.0 . Then since e -x ( )n-r
= ( )-1 -( )n-r e -x, we obtain the
desired expression immediately.
Example1 The 3rd order derivative of e -x/x
xe-x ( )3
= ( )-1 -3e-xΣr=0
3
3
r ( )1( )1+r
x-1-r
= -e -x 3
0 ( )1( )1
x -1+ 3
1 ( )1( )2
x -2+ 3
2 ( )1( )3
x -3+ 3
3 ( )1( )4
x -4
= -x
e-x
1 + x3
+ x2
6 +
x3
6
Example1' The 3rd order derivative of e -x x7
e-x x7 ( )3 = e-x
Σr=0
7
( )-1 -( )3-r 3
r ( )8-r( )8
x7- r
= e -x - 3
0 8( )8
x 7+ 3
1 7( )8
x 6- 3
2 6( )8
x 5+ 3
3 5( )8
x 4
= e-xx4 -x3 + 21 x2 - 126 x1 + 210
18.5.2 Higher Derivative of e-x log x
Formula 18.5.2
e-x log x( )n
= ex
( )-1 - n
log x -Σr=1
n
n
r xr
( )r(2.1)
Proof
Let f( )x = e-x , g( )x = log x . Then
- 15 -
e-x ( )n-r = ( )-1 - n+re-x
( )log x ( )r = ( )-1 r-1
xr
( )rr = 1,2,3,
Substituting these for Theorem 18.1.1 ,
e-x log x( )n
= Σr=0
n
n
r( )-1 - n+re-x( )log x ( )r
= n
0( )-1 - ne-x( )log x ( )0 +Σ
r=1
n
n
r( )-1 - n+re-x( )log x ( )r
= ex
( )-1 - n
log x -Σr=1
n
n
r xr
( )r
Example The 4th order derivative of e -xlog x
ex log x( )4
= ex
( )-1 - 4
log x -Σr=1
4
4
r xr
( )r
= e x
log x -
e x
1
4
1 x 1
( )1+
4
2 x 2
( )2+
4
3 x 3
( )3+
4
r x 4
( )4
= e x
log x -
e x
1
x1
4 +
x2
6 +
x3
8 +
x4
6
18.5.3 Higher Derivatives of e-x sinx , e-x cosx
Formula 18.5.3
e-xsin x( )n
= -sin 4 -n
e-xsin x-4
n(3.0s)
e-xcosx( )n
= -sin 4 -n
e-xcos x-4
n(3.0c)
Proof
Replacing x with -x in Formula 18.4.3 , we obtain the desired expressions.
Example
e-xsin x( )2
= -sin 4 -2
e-xsin x-4
2 = -2e-xcosx
e-xcosx( )3
= -sin 4 -3
e-xcos x-4
3 = -2e-x( )sin x- cosx
Higher Derivatives of e -x sinh x , e -x cosh x end now. There is no necessity for Theorem 18.1.1.
Daring use Theorem 18.1.1, we obtain the following expression first.
e-x sin x( )n
= e-xΣr=0
n
( )-1 n-r n
rsin x+
2r
- 16 -
And from this and (3.0s) , we obtain
Σr=0
n
( )-1 -r n
rsin x+
2r
= sin 4 -n
sin x-4
n
A similar expression is obtained about e-xcosx too. Then removing sin2r
, cos2r
from these, we obtain
the completely same results as Formula 18.4.3" .
18.5.4 Higher Derivatives of e-x sinhx , e-x coshx
Formula 18.5.4
e-x sinh x( )n
= e-xΣr=0
n
( )-1 -n+ r n
r 2ex - ( )-1 -re-x
(4.0s)
e-x cosh x( )n
= e-xΣr=0
n
( )-1 -n+ r n
r 2ex + ( )-1 -re-x
(4.0c)
Proof
Substituting f( )x = e -x , g( )x = sinh x , cosh x for Theorem 18.1.1 , we obtain the dsired expressions.
Example
e-x sinh x( )0
= e-xΣr=0
0
( )-1 -0+ r 0
r 2ex - ( )-1 -re-x
= e-xsinh x
e-x cosh x( )3
= e-xΣr=0
3
( )-1 -3+ r 3
r 2ex + ( )-1 -re-x
= e-x - 3
0cosh x+
3
1sinh x-
3
2cosh x+
3
3sinh x
= -4e-x( )cosh x- sinh x
Note
The following formula is known for a natural number n .
e-x sinh x( )n
= e-x cosh x( )n
= ( )-2 n-1ex( )cosh x-sinh xHowever, this formula does not hold for n =0 . That is, in this formula, the natural number n is inextensible
to the real number p . So, this is insufficient as a general formula.
- 17 -
18.6 Higher Derivatives of sin x f (x), cos x f (x)
18.6.1 Higher Derivatives of sin2x , cos2x
Formula 18.6.1
sin 2x( )n
= -2n-1cos 2x+2
n(1.0s)
cos2x( )n
= 2n-1cos 2x+2
n(1.0c)
Proof From Formula 18.6.1' mentioned next ,
cos2x( )n
= Σr=0
n
n
rcos x+
2( )n -r
cos x+2r
Here
cosA cosB = 21
1 cos( )A+B +cos( )A-B
Using this,
cos2x( )n
= 21
1Σr=0
n
n
r cos 2x+2
n + cos 2
n-r
= 21
1cos 2x+
2n
Σr=0
n
n
r +
21
1cos 2
nΣr=0
n
( )-1 r n
rAnd since
Σr=0
n
n
r = 2n , Σ
r=0
n
( )-1 r n
r = 0
substituting these for the above, we obtain (1.0c). (1.0s) is also obtained in a similar way.
Example
sin 2x( )2
= -22-1cos 2x+2
2 = 2 cos2x = 2 cos2x- sin 2x
cos2x( )3
= 23-1cos 2x+2
3 = 4 sin 2x = 8 sin x cosx
Formula 18.6.1'
sin 2x( )n
= Σr=0
n
n
rsin x+
2( )n-r
sin x+2r
(1.1s)
cos2x( )n
= Σr=0
n
n
rcos x+
2( )n -r
cos x+2r
(1.1c)
Proof
Substituting f( )x = g( )x = sin x for Theorem 18.1.1 , we obtain (1.1s). (1.1c) is also obtained in a similar
way.
- 18 -
Formula 18.6.1"
When denotes the floor function, the following expressions hold.
Σr=0
n /2
n
2r = 2n-1
(1.2e)
Σr=0
( )n -1 /2
n
2r+1 = 2n-1
(1.2o)
Proof (1.1s) is transformed as follows.
sin 2x( )n
= Σr=0
n
n
rsin x+
2( )n -r
sin x+2r
= Σr=0
n /2
n
2rsin x+
2( )n -2r
sin x+2
2r
+ Σr=0
( )n -1 /2
n
2r+1sin x+
2( )n -2r-1
sin x+2
( )2r+1
= Σr=0
n /2( )-1 r
n
2rsin x+
2( )n -2r
sin x
+ Σr=0
( )n -1 /2( )-1 r
n
2r+1sin x+
2( )n -2r-1
cosx
i.e.
sin 2x( )n
= sin x+2
nsin xΣ
r=0
n /2
n
2r - cos x+
2n
cosx Σr=0
( )n -1 /2
n
2r+1On the other hand, (1.0s) is transformed as follows too.
sin 2x( )n
= 2n-1sin x+2
nsin x - 2n-1cos x+
2n
cosx
From these, the following expression follows .
sin x +2
nsin x Σ
r=0
n /2
n
2r- 2n-1 = cos x +
2n
cos x Σr=0
( )n -1 /2
n
2r+1- 2n-1
In order to hold this equation for arbitrary x , the followings are necessary.
Σr=0
n /2
n
2r- 2n-1 = 0 , Σ
r=0
( )n -1 /2
n
2r+1- 2n-1 = 0
In addition, this formula is known. (See "岩波 数学公式Ⅱ" p11)
18.6.2 Higher Derivatives of sin3x , cos3x
Formula 18.6.2
sin 3x( )n
= 43
sin x+2
n - 4
3n
sin 3x+2
n(2.0s)
cos3x( )n
= 43
cos x+2
n + 4
3n
cos 3x+2
n(2.0c)
- 19 -
Proof From Formula 18.6.2' mentioned next, it is obtained in a similar way in the case of the 2nd degree.
However, it is not so easy as the case of the 2nd degree. ( See 20.1.3 )
Example
sin 3x( )2
= 43
sin x+2
2 - 4
32
sin 3x+2
2 = - 4
3sin x + 4
9sin 3x
cos3x( )3
= 43
cos x+2
3 + 4
33
cos 3x+2
3 = 4
3sin x + 4
27sin 3x
Formula 18.6.2'
sin 3x( )n
= sin 2x( )0
sin x+2
n
- Σr=1
n
n
r2r-1cos 2x+
2r
sin x+2
( )n -r (2.1s)
cos3x( )n
= cos2x( )0
cos x+2
n
+ Σr=1
n
n
r2r-1cos 2x+
2r
cos x+2
( )n -r (2.1c)
Proof
Substituting f( )x = sin 2x , g( )x = sin x for Theorem 18.1.1, we obtain (2.1s). (2.1c) is also obtained in
a similar way.
Formula 18.6.2"
Σr=0
n /222r-1
n
2r = 4
3n + ( )-1 n
(2.2e)
Σr=0
( )n -1 /222r
n
2r+1 = 4
3n - ( )-1 n
(2.2o)
Proof From (2.1s)
sin 3x( )n
= sin 2x( )0
sin x +2
n -Σ
r=1
n
n
r2r-1cos 2x +
2r
sin x +2
( )n -r
= sin 2x( )0
sin x+2
n
- Σr=0
( )n -1 /2
n
2r+122rcos 2x+
2( )2r+1
sin x+2
( )n -2r-1
- Σr=1
n /2
n
2r22r-1cos 2x+
22r
sin x+2
( )n -2r
- 20 -
= 21
- 2-1cos2x sin x+2
n
+ Σr=0
( )n -1 /2( )-1 r
n
2r+122rsin 2x sin x+
2( )n -2r-1
- Σr=1
n /2( )-1 r
n
2r22r-1cos2x sin x+
2( )n-2r
= 21
sin x+2
n - Σ
r=0
( )n -1 /2
n
2r+122rsin 2x cos x+
2n
- n
02-1cos2x sin x+
2n
- Σr=1
n /2
n
2r22r-1cos2x sin x+
2n
= 21
sin x+2
n - Σ
r=0
( )n -1 /2
n
2r+122rsin 2x cos x+
2n
- Σr=0
n /2
n
2r22r-1cos2x sin x+
2n
When n=1 ,
sin 3x( )1
= 21
sin x +2
1 - cos 2x sin x +
21
2-1 1
0- sin 2x cos x +
21
20 1
1
= 21
sin x+2
1 - 2
1cos2x cosx + 1 sin 2x sin x
= 21
sin x+2
1 + 2
1sin2x sin x - 2
1( )cos2x cosx - sin 2x sin x
= 21
sin x+2
1 + 2
1sin2x sin x - 2
1cos3x
= 21
sin x+2
1 - 2
121
( )cos3x - cosx - 21
cos3x
= 43
sin x+2
1 - 4
31
sin 3x+2
1
Σr=0
1/222r-1
1
2r = 4
31-1 = 2
1 = 4
31+( )-1 1
Σr=0
( )1-1 /222r
1
2r+1 = 4
31-1+
21
= 431+1
= 1 = 431-( )-1 1
When n=2 ,
sin 3x( )2
= 21
sin x+2
2 - cos2x sin x+
22 2-1
2
0+ 21
2
2
- sin2x cos x+2
220
2
1
- 21 -
= 21
sin x+2
2 + 2
5cos2x sin x + 2 sin 2x cosx
= 21
sin x+2
2 + 2
121
( )sin 3x - sin x + 2sin 3x
= 43
sin x+2
2 + 4
9sin 3x
= 43
sin x+2
2 - 4
32
sin 3x+2
2
Σr=0
2/222r-1
2
2r = 4
32-1+
21
= 25
= 432+1
= 432+( )-1 2
Σr=0
( )2-1 /222r
2
2r+1 = 4
32-1 = 2 = 4
32-( )-1 2
Hereafter, by induction, we obtain the desired expressions.
If both sides of Formula18.6.2" are illustrated, it is as follows. The left side is blue line and the right side is
red point.
- 22 -
18.6.3 Higher Derivatives of the product of trigonometric and hyperbolic functions
Formula 18.6.3
( )sinxsinhx( )n
= Σr=0
n
n
rsin x+
2( )n -r
2ex -( )-1 -re-x
(3.1)
( )sinxcoshx( )n
= Σr=0
n
n
rsin x+
2( )n -r
2ex +( )-1 -re-x
(3.2)
( )cosxsinhx( )n
= Σr=0
n
n
rcos x+
2( )n -r
2ex -( )-1 -re-x
(3.3)
( )cosxcoshx( )n = Σ
r=0
n
n
rcos x+
2( )n -r
2ex +( )-1 -re-x
(3.4)
Proof
Substituting f( )x = sin x , g( )x = sinh x for Theorem 18.1.1, we obtain (3.1).
The others are also obtained in a similar way.
Example
( )sinxcoshx( )0
= Σr=0
0
0
rsin x+
2( )0-r
2ex +( )-1 -0e-x
= sinxcoshx
( )cosxcoshx( )2
= Σr=0
2
2
rcos x+
2( )2-r
2ex +( )-1 -re-x
= cos x+2
2coshx +2cos x+
21
sinhx +cos x+2
0coshx
= -cosxcoshx - 2sin xsinhx + cosxcoshx = -2sin xsinhx
- 23 -
18.7 Higher Derivatives of sinh x f (x), cosh x f (x)
18.7.1 Higher Derivatives of sinh2x , cosh2x
Formula 18.7.1
sinh 2x( )n
= Σr=0
n
n
r 2ex -( )-1 -n+re-x
2ex -( )-1 -re-x
(1.1s)
cosh 2x( )n
= Σr=0
n
n
r 2ex +( )-1 -n+re-x
2ex +( )-1 -re-x
(1.1c)
Proof
Substituting f( )x = g( )x = sinh x for Theorem 18.1.1, we obtain (1.1s). (1.1c) is also obtained in
a similar way.
Example
sinh 2x( )0
= Σr=0
0
0
r 2ex -( )-1 -0+re-x
2ex -( )-1 -re-x
= sinh 2x
cosh 2x( )3
= Σr=0
3
3
r 2ex +( )-1 -3+re-x
2ex +( )-1 -re-x
= 3
0sinhxcoshx +
3
1coshxsinhx +
3
2sinhxcoshx +
3
3coshxsinhx
= 8sinhxcoshx = 4sinh( )2x
2007.05.06
K. Kono
Alien's Mathematics
- 24 -