19 Super Derivative of the Product of Two Functions

19.1 Super Leibniz Rule

Theorem 19.1.1

Let ( )x,y be the beta function and p be a positive number. And for r =0,1,2, , let f < >-p+ r be

arbitrary primitive function of f( )x and g( )rbe the r th order derivative function of g( )x .

Then the following expressions hold.

f( )x g( )x ( )p = Σr=0

m -1

p

rf ( )p-r ( )x g( )r ( )x + Rm

p(1.1)

Rmp = ( )-p ,m

( )-1 m

Σk=0

m+ k1

-p -1

k f< >m+ k ( )x g( )m+ k ( )x

( )p(1.1r)

Especially, when n =0,1,2,

f( )x g( )x ( )n = Σr=0

n

n

rf ( )n-r ( )x g( )r ( )x ( Leibniz ) (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

When n =1 ∑∑∑ of the 3rd line does not exist, when n =0 ∑∑ of the 2nd line does not exist also. And the

upper limit n-1 of ∑ of the 4th line can be replaced by . Therefore, if the index n of the integration

operator is substituted for -n in consideration of these, it is 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

Analytically continuing the index of the integration operator to [ ]0 ,p from[ ]1 ,n ,

a( )p

x

a( )0

x

f < >0 g( )0 dx-p = Σr=0

m -1

p

rf < >-p+ r g( )r

+ ( )-p ,m( )-1 m

Σk=0

m+k1

-p -1

k a( )p

x

a( )0

x

f< >m+ k g( )m+ k dx-p

- 1 -

Then, replacing the integration operators dx -p,< >-p +r with the differentiation operators ( )p ,( )p -rrespectivly, we obtain (1.1) and (1.1r) .

Especially, when p = m-1 , m=1,2,3, , since ( )-p ,m = ( )1-m, m = , Rmp =0 .

Then

f( )x g( )x ( )m-1 = Σr=0

m -1

m-1

rf( )m-1-r g( )r

Furthermore, replacing m-1 with n , we obtain (1.1') .

Q.E.D

- 2 -

19.2 Super Derivative of x̂ a f (x)

Formula 19.2.0

Let ( )z be the gamma function, ( )n ,m be the beta function, f< >r be arbitrary the r th primitive

function of f( )x and f < >ra be the function values of f< >r

on a . Then the following expressions hold for

a positive number p .

(1)

x f( )x( )p

= Σr=0

m -1

p

r ( )1+-r( )1+

x- rf ( )p-r

+ ( )-p ,m( )-1 m

Σk=0

m+k1

-p -1

k ( )1+-m-k( )1+

x- m- kf< >m+ k( )p

(0.1)

Especially, when m=0,1,2,

xm f( )x( )p

= Σr=0

m

p

r ( )1+m-r( )1+m

xm- rf ( )p-r(0.1')

Where, if = -1,-2,-3, , it shall read as follows.

1+-r

( )1+ ( )-1 -r

( )-( )-+r

(2) When -1,-2,-3, & -p -1,-2,-3,

xm f( )x( )p

= Σr=0

m -1

p

r ( )1+-p +r( )1+

x-p+ rf( )r

+ ( )-p ,m( )-1 m

Σk=0

m+k1

-p -1

k ( )1++m+k( )1+

x+m+kf( )m+ k( )p

(0.2)

Proof

Since differentiation is an inverse operation of integration, replacing the index p of the integration operator

with -p in Formula17.2.0 in 17.2 , we obtain the desired expressions.

19.2.1 Super Derivative of ( )ax+b p( )cx+d q

Formula 19.2.1

The following expressions hold for p >0 , s >0 such that p -s-1,-2,-3,

( )ax+b p( )cx+d q ( )s

= Σr=0

m -1

s

r ( )1/c r

( )1/a -s+ r

( )1+p -s+r ( )1+q -r( )1+p ( )1+q

( )cx+d r-q

( )ax+b p-s+r

+ Rms

(1.1)

Rms = ( )-s,m

( )-1 m

Σk=0

m+ k1

-s-1

k ac m+ k

( )1+p +m+k ( )1+q -m-k( )1+p ( )1+q

( )ax+b p+m+ k( )cx+d q-m- k ( )s

(1.1r)

limm

Rms = 0

- 3 -

Especially, when m=0,1,2,

( )ax+b p ( )cx+d m ( )s

= Σr=0

m

s

r ( )1/c r

( )1/a -s+ r

( )1+p -s+r ( )1+m-r( )1+p ( )1+m

( )cx+d r-m

( )ax+b p-s+ r

(1.1')

Proof

Although it is an original way to substitute f( )x =( )ax +b p , g( )x =( )cx +d q for Theorem 19.1.1 ,

here, we reverse the sign of the index of the integration operator < >s in Formula 17.3.1 ( 17.3 ) and replace

it with the differentiation operator ( )s .

Example1 The 1/2th order derivative of x-2 3 3x+4

Substituting a =1, b =-2, p=1/2 , c=3, d =4, q =1/3 , s=1/2 for (1.1) ,

x-2 3 3x+4 2

1

= Σr=0

1/2

r3r

( )1+r ( )4/3-r( )3/2 ( )4/3

( )x-2 r( )3x+4 31

- r

The left side is calculated by the expression which replaced the order of integration and differentiation in

Riemann-Liouville differintegral. The integration lower limit is taken as -b /a = 2 according to Example1

in 17.3 . When m=17, the values of the both sides on arbitrary point x =5 are as follows.

Example1' The 1/2th order derivative of 3 3x+4( )x-2 2

Substituting a =3, b =4, p =1/3 , c=1, d =-2, m=2 , s=1/2 for (1.1') ,

3 3x+4( )x-2 2 21

= Σr=0

2

1/2

r3 2

1- r

( )5/6+r ( )3-r( )4/3 ( )3

( )x-2 r-2

( )3x+4- 6

1+ r

The integration lower limit of Riemann-Liouville differintegral in the left side is taken as -b /a = -4/3 according to Example1' in 17.3.1 . The values of the both sides on arbitrary point x =3 are as follows.

Example2 The 1/2th order derivative of x-2 /( )3x+4 When q = -1,-2,-3,

- 4 -

1+q -r

( )1+q = ( )-1 -r

( )-q( )-q +r

Then (1.1) can be read as follows.

( )ax+b p( )cx+d q ( )s

= Σr=0

m -1

s

r ( )-1/c r

( )1/a -s+ r

( )1+p -s+r ( )-q( )1+p ( )-q +r

( )cx+d r-q

( )ax+b p-s+ r

+ Rms

Substituting a =1, b=-2, p =1/2 , c=3, d =4, q =-1 , s=1/2 for this ,

3x+4x-2

21

=Σr=0

1/2

r( )-3 r

( )1+r ( )1( )3/2 ( )1+r

( )3x+4 r+1

( )x-2 r

The integration lower limit of Riemann-Liouville differintegral in the left side is -b /a = 2 like Example1 .

When m=10, the values of the both sides on arbitrary point x =4 are as follows.

19.2.2 Super Derivative of x log x

Formula 19.2.2

When ( )z , ( )z denotes the gamma function and the digamma function respectively, the following

expressions hold.

(1)

x log x( )p

= Σr=0

m -1

p

r ( )1-p+rlog x -( )1-p+r -

( )1+-r( )1+

x -p + Rmp

(2.1)

Rmp =

( )-p ,m( )-1 m

Σk=0

m+ k1

-p -1

k ( )1+m+klog x -( )1+m+k -

( )1+-m-k( )1+ x ( )p

(2.1r)

limm

Rmp = 0

Especially, when m=0,1,2,

xmlog x( )p

= Σr=0

m

p

r ( )1-p+rlog x-( )1-p+r -

( )1+m-r( )1+m

xm- p(2.1')

(2) When -1,-2,-3, & -p -1,-2,-3,

x log x( )p

= 1+-p( )1+

x-p log x

+ Σr=1

m -1

( )-1 r-1 p

r 1+-p +r

( )1+ ( )rx-p + Rm

p(2.2)

- 5 -

Rmp = ( )-p ,m

x-p

1+-p( )1+

Σk=0

( )-1 k-1

-p -1

k m+ k( )1+ , m+ k

(2.2r)

limm

Rmp = 0

Proof

Substituting f( )x = log x for Formula 19.2.0 , we obtain the desired expressions.

Example1 The 4/5th order derivative of x3/2 log x Substituting =3/2 , p=4/5 for (2.1) ,

x 23

log x

( )4/5

= Σr=0

m -1

4/5

r ( )1/5 +rlog x -( )1/5 +r -

( )5/2-r

( )5/2x 10

7

+ Rm54

The integration lower limit of Riemann-Liouville differintegral in the left side is x =0 .

When m=120, the values of both sides on arbitrary point x =3 are as follows.

Example1' The 1.3th order derivative of x3log x Substituting m=3 , p =1.3 for (2.1') ,

x3log x( )1.3

= Σr=0

3

1.3

r ( )-0.3+rlog x-( )-0.3+r -

( )4-r( )4

x1.7

The values of the both sides on arbitrary point x=2.1 are as follows.

The remainder in the super derivative of the product of two functions becomes a series of super derivatives.

So it is very difficult to calculate this. However, the calculation of the above (2.2r) is exceptionally easy.

So we show it as follows.

- 6 -

Example2 The 1/2th order derivative of 3

x log x Substituting =1/3 , p=1/2 for the above (2) ,

x log x( )p

= 5/6( )4/3

x- 6

1 log x

+ Σr=1

m -1

( )-1 r-1 -1/2

r 5/6+r

( )4/3 ( )rx

- 61

+ Rm21

Rm21

= ( )-1/2,mx

- 61

5/6( )4/3

Σk=0

( )-1 k-1

-3/2

k m+ k( )4/3, m+ k

When m=10, the values of both sides on arbitrary point x =1 are as follows.

Complete Automorphism

Reversing the sign of the index of the integration operator < >p in Formula 17.3.2 in 17.3 , we obtain the

following expression without the remainder term. However, it is complicated and the convergence is also very

slow.

Formula 19.2.2'

x log x( )p

=

1 - pΣk=0

-p -1

k ( )1+ k 2( )1+k ,-k

1

( )1-plog x -( )1-p -

- p 1+-p( )1+

Σk=0

-p -1

k ( )1+ k 2( )1+k ,-k

( )2+k +

x -p

19.2.3 Super Derivatives of x sinx , x cosx

Formula 19.2.3

When m=0,1,2, , the following expressions hold for p >0 .

- 7 -

xm sin x( )p

=Σr=0

m

p

r ( )1+m-r( )1+m

xm- rsin x+2

( )p -r (3.1s)

xm cosx( )p

=Σr=0

m

p

r ( )1+m-r( )1+m

xm- rcos x+2

( )p -r (3.1c)

Proof

Reversing the sign of the index of the integration operator < >p in Formula 17.3.3 in 17.3 and replacing it

with the differentiation operator ( )p , we obtain the desired expressions.

Example The 3/2th order derivative of x3sinx Substituting m=2 , p =3/2 for (3.1s) ,

x3 sin x 2

3

=Σr=0

3

3/2

r ( )4-r( )4

x3-r sin x+2

( )3/2-r

This super derivative can not be examined by Riemann-Liouville differntegral. Then we draw this with the 1st and

the 2nd order derivative on a figure side by side. It is as follows.

Formula 19.2.3' (Collateral Super Derivative)

The following expressions hold for ,p such that -1,-2,-3, & - p-1,-2,-3, .

x sin x{ }p

=Σr=0

m -1

p

r ( )1+-p +r( )1+

x-p+ rsin x+2r

+ Rmp

(3.2s)

Rmp = ( )-p , m

( )-1 m

Σk=0

m + k1

-p-1

k 1+ +m +k( )1+

x +m+ k sin x +2

m + k ( )p

x cosx{ }p

=Σr=0

m -1

p

r ( )1+-p +r( )1+

x-p+ rcos x+2r

+ Rmp

(3.2c)

Rmp = ( )-p , m

( )-1 m

Σk=0

m + k1

-p-1

k 1+ +m +k( )1+

x +m+ k cos x +2

m + k ( )p

limm

Rmp = 0

- 8 -

Proof

Reversing the sign of the index of the integration operator < >p in Formula 17.3.3' in 17.3 and replacing it

with the differentiation operator ( )p , we obtain the desired expressions.

Example Collateral the 1/2th order derivative of x3/2 sinx Substituting =3/2 , p=1/2 for (3.2s), we obtain

x 23

sin x

21

=Σr=0

m -1

1/2

r ( )2-r( )5/2

x1-r sin x+2r

+ Rm21

The integration lower limit of Riemann-Liouville differintegral in the left side is x =0 .

When m=15, the values of the both sides on arbitrary point x =5 are as follows.

19.2.4 Super Derivatives of x sinhx , x coshx

Formula 19.2.4

When m=0,1,2, , the following expressions hold for p >0 .

xm sinh x{ }p

= Σr=0

m

p

r ( )1+m-r( )1+m

xm- r

2ex -( )-1 -p+ re-x

(4.1s)

xm cosh x{ }p

= Σr=0

m

p

r ( )1+m-r( )1+m

xm- r

2ex +( )-1 -p+ re-x

(4.1c)

Example The 0.999th order derivaive of x3 coshx Substituting m=3 , p =0.999 for (4.1c), we obtain

x3cosh x{ }0.999

= Σr=0

3

0.999

r ( )4-r( )4

x3- r

2ex +( )-1 -0.999+ re-x

This super derivative can not be examined by Riemann-Liouville differntegral. Then we calculate the values on

arbitrary point x=2 for the 0.999th order derivative and the 1st order derivative respectively. Although the former

is a complex number, the real part is naturally near to the coefficient of the later.

- 9 -

Formula 19.2.4' (Collateral Super Derivative)

The following expressions hold for ,p such that -1,-2,-3, & - p-1,-2,-3, .

x sinh x{ }p

=Σr=0

m -1

p

r 1+-p +r( )1+

x-p+ r

2ex-( )-1 -re-x

+ Rmp

(4.2s)

Rmp = ( )-p , m

( )-1 m

Σk=0

m + k1

-p-1

k 1+ +m + k( )1+

x +m+ k

2e x-( )-1 -m-ke -x ( )p

x cosh x{ }p

=Σr=0

m -1

p

r 1+-p +r( )1+

x-p+ r

2ex+( )-1 -re-x

+ Rmp

(4.2c)

Rmp = ( )-p , m

( )-1 m

Σk=0

m + k1

-p-1

k 1+ +m + k( )1+

x +m+ k

2e x+( )-1 -m-ke -x ( )p

limm

Rmn = 0

Example Collateral the 3/2th order integral of 3

x sinhx Substituting =1/3 , p=3/2 for (4.2s), we obtain

3x sinh x

23

= Σr=0

m -1

3/2

r -1/6+r( )4/3

x- 6

1+ r

2ex -( )-1 -re-x

+ Rm3/2

The integration lower limit of Riemann-Liouville differintegral in the left side is x =0 .

When m=50, the values of the both sides on arbitrary point x =17 are as follows.

- 10 -

19.3 Super Derivative of log x f (x)

19.3.1 Super Derivative of ( )logx 2

Formula 19.3.1

log 2x( )p

= 1-px-p log x log x- 1-p -

- x-p Σr=1

m -1

( )-1 r p

r 1-p +r log x-( )1-p+r - ( )r

+ Rmp

(1.1)

Rmp = ( )-p ,m 1-p

x -p

Σk=0

( )m+ k 2

( )-1 k-1

-p -1

k log x - 1-p -( )1+m+k -2 (1.1r)

limm

Rmp = 0

Proof

Reversing the sign of the index of the integration operator < >p in Formula 17.4.1 in 17.4 and replacing it

with the differentiation operator ( )p , we obtain the desired expressions.

Example The 1/2th order derivative of ( )logx 2

When m=4000, the values of both sides on arbitrary point x =2.7 are as follows. Since the convergence

is slow, even if it calculates so far, both sides does not match only up to 3 digits after the decimal point.

- 11 -

19.4 Super Derivative of e x̂ f (x)

19.4.1 Super Derivative of ex x

Formula 19.4.1

ex x( )p

= exΣr=0

m -1

p

r 1+-r( )1+

x- r + Rmp

(1.1)

Rmp = ( )-p ,m

( )-1 m

Σk=0

m+ k1

-p-1

k 1+-m-k( )1+

ex x- m-k ( )p(1.1r)

Especially, when m=0,1,2,

ex xm ( )p = exΣ

r=0

m

p

r ( )1+m-r( )1+m

xm- r(1.1')

Proof

Let f( )x = ex , g( )x = x . Then

x ( )r = 1+-r

( )1+ x- r ( )-1,-2,-3,

Substituting these for Theorem 19.1.1 , we obtain (1.1) and (1.1r).

Especially, when =m-1 m =1,2,3, , ( )1+-m-k = k =0,1,2,3, .

Then, the remainder term disappears and (1.1) is as follows.

ex xm-1 ( )p = exΣ

r=0

m -1

p

r 1+m-1-r( )1+m-1

xm-1- r

Thus, replacing m-1 with m , we obtain (1.1').

Example1 The 1/2.1th order integral of ex x3/4

Substituting =3/4 , p=1/2 for (1.1) ,

ex x3/4 21

= exΣr=0

m -1

1/2

r 1+3/4-r( )1+3/4

x 43

- r + Rm

1/2

The left side is calculated by the expression which replaced an order of integration and differentiation in

Riemann-Liouville differintegral. When m=10, the values of the both sides on arbitrary point x =10 are

as follows.

Example1' The 5/2th order derivative of ex x7

Substituting m=7 , p =5/2 for (1.1') ,

- 12 -

ex x7 25

= exΣr=0

7

5/2

r 8-r( )8

x7- r

The values of the both sides on arbitrary point x =.5 are as follows.

Formula 19.4.1" ( Collateral Super Derivative)

The following expression holds for ,p such that -1,-2,-3, & - p-1,-2,-3, .

ex x( )p

= ex Σr=0

m -1

p

r 1+-p +r

( )1+x-p+ r + Rm

p(1.1")

Rmp = ( )-p ,m

( )-1 m

Σk=0

m+ k1

-p-1

k 1++m+k( )1+

ex x+ m+k ( )p

limm

Rmp = 0

Example1" Collateral the 1/2th order derivative of ex x3/4

Substituting =3/4 , p=1/2 for (1.1") , we obtain

ex x3/4 21

= ex Σr=0

m -1

1/2

r 3/4+r

( )3/2x 4

3+ r

+ Rm1/2

The values of both sides on the same point x =-10 as Example1 are as follows. Though the both sides are

corresponding, they are considerably different from the values of Example1 (Lineal the 1/2th order derivative.)

19.4.2 Super derivative of ex log x

All the polynomials obtained by applying Theorem 19.1.1 to exlog x become asymptotic expansions, and

they are hardly helpful.

19.4.3 Super derivatives of ex sinx , ex cosx

- 13 -

Formula 19.4.3

exsin x( )p

= sin 4 -p

exsin x+4

p(3.0s)

excosx( )p

= sin 4 -p

excos x+4

p(3.0c)

Proof

Analytically continuing the index of the differentiation operator in Formula 18.4.3 in 18.4 to [ ]0 ,p from

[ ]1 ,n , we obtain the desired expressions.

Example The 1/3th order derivative of ex sinx When p =1/3 , the values of the both sides of (3.0s) on arbitrary point x =1.2 are as follows.

Trigonometric Series If Formula 18.4.3' in 18.4 is extended to the real number, the following trigonometric series are obtained.

Formula 19.4.3'

Σr=0

p

rsin x+

2r

= sin 4 -p

sin x+4

p(3.1s)

Σr=0

p

rcos x+

2r

= sin 4 -p

cos x+4

p(3.1c)

Especially, when x=0 ,

Σr=0

p

rsin 2

r = sin 4

-p

sin 4p

(3.1's)

Σr=0

n

p

rcos 2

r = sin 4

-p

cos 4p

(3.1'c)

Alternating Binomial Series

Removing sin2r

, cos2r

from (3.1's), (3.1'c), we obtain the following interesting series.

Formula 19.4.3"

Σk=0

( )-1 k

p

2k+1 = 2 2

p

sin 4p

(3.2s)

- 14 -

Σk=0

( )-1 k

p

2k= 2 2

p

cos 4p

(3.2c)

Proof Since the odd-numbered terms of the left side in (3.1's) are all 0,

Σr=0

p

rsin 2

r =

p

1-

p

3+

p

5-

p

7+- = Σ

k=0

( )-1 k

p

2k+1

Σk=0

( )-1 k

p

2k+1 = sin 4

-p

sin 4p

= 2 2p

sin 4p

Next, since the even-numbered terms of the left side in (3.1'c) are all 0,

Σr=0

n

p

rcos 2

r =

p

0-

p

2+

p

4-

p

6+- = Σ

k=0

( )-1 k

p

2k

Σk=0

( )-1 k

p

2k = sin 4

-p

cos 4p

= 2 2p

cos 4p

If a horizontal axis is set as p and these are illustrated , it is as follows. Although the left side is blue and

right side is red, since both sides overlap exactly, the left side (blue) is not visible.

19.4.4 Super Derivatives of ex sinhx , ex coshx

Formula 19.4.4

exsinh x( )p

= exΣr=0

p

r 2ex - ( )-1 -re-x

(4.0s)

excosh x( )p

= exΣr=0

p

r 2ex + ( )-1 -re-x

(4.0c)

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Example The 3/2th order derivative of ex coshx When p =3/2 , the values of the both sides of (4.0c) on arbitrary point x =1.3 are as follows.

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19.5 Super Derivative of f (x) / e^x

19.5.1 Super Derivative of e-x x

Formula 19.5.1

e-x x( )p

= ex

( )-1 -p

Σr=0

m -1

( )-1 r p

r 1+-r( )1+

x- r + Rmp

(1.1)

Rmp = ( )-p ,m

1Σk=0

m+ k( )-1 k

-p -1

k 1+-m-k( )1+

e x

x - m-k ( )p

(1.1r)

Especially, when m=0,1,2,

e-x xm ( )p =

ex

( )-1 -p

Σr=0

m

( )-1 r p

r 1+m-r( )1+m

xm- r(1.1')

Example1 The 1/2th order derivative of e-x x If we substitute =1/2 , p =1/2 for (1.1) and calculate the values of the both sides on arbitrary point

x =10 , it is as follows. In addition, this is an asymptotic expansion.

Example1' The 5/2th order derivative of e-x x7

If we substitute m=7 , p =5/2 for (1.1') and calculate the values of the both sides on arbitrary point

x =6 , it is as follows.

Formula 19.5.1" ( Collateral Super Derivative)

The following expression holds for ,p such that -1,-2,-3, & - p-1,-2,-3, .

e-x x( )p

= ex

1Σr=0

m -1

( )-1 - r p

r 1+-p +r

( )1+x-p+ r + Rm

p(1.1")

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Rmp = ( )-p ,m

1Σk=0

m+ k( )-1 -k

-p -1

k 1++m+k( )1+ e x

x - m-k ( )p

limm

Rmp = 0

Example1" Collateral the 1/2th order derivative of e-x x If we substitute =1/2, p =1/2, m=30 for (1.1") and calculate the values of the both sides on arbitrary

point x =10 , it is as follows. Though the both sides are corresponding, they are completely different from

the values of Example1 (Lineal the 1/2th order derivative.)

19.5.2 Super Derivative of e-x log x

All the polynomials obtained by applying Theorem 19.1.1 to e-xlog x become asymptotic expansions, and

they are hardly helpful.

19.5.3 Super derivatives of e-x sinx , e-x cosx

Formula 19.5.3

e-xsin x( )p

= ( )-1 -p sin 4 -p

e-xsin x-4

p(3.0s)

e-xcosx( )p

= ( )-1 -p sin 4 -p

e-xcos x-4

p(3.0c)

Example The 3/2th order derivative of e-x sinx The values of the both sides of (3.0s) on arbitrary point x =1.7 are as follows.

19.5.4 Super derivatives of e-x sinhx , e-x coshx

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Formula 19.5.4

e-x sinh x( )p

= e-xΣr=0

( )-1 -p+ r

p

r 2ex - ( )-1 -re-x

(4.0s)

e-x cosh x( )p

= e-xΣr=0

( )-1 -p+ r

p

r 2ex + ( )-1 -re-x

(4.0c)

Example The 3/2th order derivative of e-x coshx The values of the both sides of (4.0c) on arbitrary point x =1.3 are as follows.

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19.6 Super Derivatives of sin x f (x), cos x f (x)

19.6.1 Super Derivatives of sin2x , cos2x

Formula 19.6.1

sin 2x( )p

= -2p-1cos 2x+2

p(1.0s)

cos2x( )p

= 2p-1cos 2x+2

p(1.0c)

Proof

Analytically continuing the index of the differentiation operator in Formula 18.6.1 in 18.6.1 to [ ]0 ,p from

[ ]1 ,n , we obtain the desired expressions.

Example

sin 2x 2

1

= -2 21

-1cos 2x+

4

= 21

( )sin 2x+ cos2x

cos2x 2

3

= 2 23

-1cos 2x+

43

= -( )sin 2x+ cos2x

Analytically continuing the index of the differentiation operator in Formula 18.6.1' and Formula 18.6.1" in 18.6

to [ ]0 ,p from[ ]1 ,n respectively, we obtain the following two formualas.

Formula 19.6.1'

sin 2x( )p

= Σr=0

p

rsin x+

2( )p -r

sin x+2r

(1.1s)

cos2x( )p

= Σr=0

p

rcos x+

2( )p -r

cos x+2r

(1.1c)

Formula 19.6.1"

Σr=0

p

2r = 2p-1 p >0 (1.2e)

Σr=0

p

2r+1 = 2p-1 p >0 (1.2o)

If a horizontal axis is set as p and these are illustrated , it is as follows. (1.2e) , (1.2o) and 2p-1 are blue,

red and green respectively. Since three curves overlap exactly, only red ( 2p-1) is visible.

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The following formula follows from Formula 19.6.1 immediately.

Formula 19.6.1"'

Σr=0

p

r = 2p p >0 (1.3)

Note In fact, Formula 19.6.1" and Formula 19.6.1"' hold for p >-1 .

19.6.2 Super Derivatives of sin3x , cos3x

Formula 19.6.2

sin 3x( )p

= 43

sin x+2

p - 4

3p

sin 3x+2

p(2.0s)

cos3x( )p

= 43

cos x+2

p + 4

3p

cos 3x+2

p(2.0c)

Example

sin 3x 2

1

= 43

sin x+4

- 43

sin 3x+4

cos3x 2

3

= 43

cos x+4

3 + 4

3 3cos 3x+

43

If the later is drawn with the 1st and the 2nd order derivatives on a figure side by side, it is as follows.

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19.6.3 Super Derivatives of the product of trigonometric and hyperbolic functions

Formula 19.6.3

( )sinxsinhx( )p

= Σr=0

p

rsin x+

2( )p -r

2ex -( )-1 -re-x

(3.1)

( )sinxcoshx( )p

= Σr=0

p

rsin x+

2( )p -r

2ex +( )-1 -re-x

(3.2)

( )cosxsinhx( )p

= Σr=0

p

rcos x+

2( )p-r

2ex -( )-1 -re-x

(3.3)

( )cosxcoshx( )p

= Σr=0

p

rcos x+

2( )p-r

2ex +( )-1 -re-x

(3.4)

Example The 3/2th order derivative of cosxsinhx

( )cosxsinhx( )3/2

= Σr=0

3/2

rcos x+

2( )3/2-r

2ex -( )-1 -re-x

If this is drawn with the 1st and the 2nd order derivatives on a figure side by side, it is as follows.

1 2 3 4

-10

-5

5

10

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19.7 Super Derivatives of sinh x f (x), cosh x f (x)

19.7.1 Super Derivatives of sinh2x , cosh2x

Formula 19.7.1

sinh 2x( )p

= Σr=0

p

r 2ex -( )-1 -p+re-x

2ex -( )-1 -re-x

(1.1s)

cosh 2x( )p

= Σr=0

p

r 2ex +( )-1 -p+re-x

2ex +( )-1 -re-x

(1.1c)

Proof

Analytically continuing the index of the differentiation operator in Formula 18.7.1 in 18.7 to [ ]0 ,p from

[ ]1 ,n , we obtain the desired expressions.

Example: The 0.01th and the 0.99th order derivatives of sinh2x According to (1.1s), if each differential coefficient of the 0th order, the 0.01th order, the 0.99th order and

the 1st order on arbitrary point x =2 are calculated , they are as follows. Naturally, the super differential

coefficients turn into complex numbers.

2010.11.11

K. Kono

Alien's Mathematics

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