Derivative of a FunctionDate: 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/ |
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