Derivative of a Function

Date: 04/01/99 at 02:13:28
From: Steve Yang
Subject: Formula for calculating Kth order derivative of function

I am trying to find a formula for calculating the Kth order derivative 
of the function: f(x) = exp(A(x)). I hope you can help me solve this 
question.

Thanks a lot!

- Steve

Date: 04/01/99 at 09:58:36
From: Doctor Rob
Subject: Re: Formula for calculating Kth order derivative of function

Thanks for writing to Ask Dr. Math.

The form for the Kth order derivative of exp(A(x)) is exp(A(x)) times
a sum. There is one term of the sum for every partition of the number
K into positive integer parts. If the partition is

   K = 1*n(1) + 2*n(2) + ... + r*n(r),

then the term has the form of a product for i from 1 to r of the
derivatives [d^iA/dx^i]^n(i), times the coefficent

   K!/([PRODUCT(i!)^n(i)]*n(i)!).

Example:  

K = 4. There are five partitions of 4 into positive integer parts:

   4 = 4,  r = 4, n(1) = 0, n(2) = 0, n(3) = 0, n(4) = 1,
   4 = 3 + 1,  r = 3, n(1) = 1, n(2) = 0, n(3) = 1,
   4 = 2 + 2 = 2*2,  r = 2, n(1) = 0, n(2) = 2,
   4 = 2 + 1 + 1 = 2 + 1*2,  r = 2, n(1) = 2, n(2) = 1,
   4 = 1 + 1 + 1 + 1 = 1*4,  r = 1, n(1) = 4.

The terms look like

   4!/[4!*1!]               * d^4A/dx^4,
   4!/[(3!*1!)*(1!*1!)]     * (d^3A/dx^3)*(dA/dx),
   4!/[(2!)^2*2!]           * (d^2A/dx^2)^2,
   4!/[(2!*1!)*([1!]^2*2!)] * (d^2A/dx^2)*(dA/dx)^2,
   4!/[(1!)^4*4!]           * (dA/dx)^4.

Then the 4th derivative is

   exp(A(x))*[d^4A/dx^4 + 4*(d^3A/dx^3)*(dA/dx) + 3*(d^2A/dx^2)^2 +
              6*(d^2A/dx^2)*(dA/dx)^2 + (dA/dx)^4].

Formally, the Kth derivative can be written

   d^K[exp(A[x])]/dx^K = exp(A[x]) *

                                  K
             SUM          K! * PRODUCT [d^iA/dx^i]^n(i)/[(i!)^n(i)*
n(i)!].
        K                        i=1
     K=SUM i*n(i), n(i)>=0
       i=1

This can be proven using the Principle of Mathematical Induction on K.
The proof is pretty complicated, but uses just elementary facts.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/