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 -