Higher Derivative of the Product of Two...

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 -