# High order derivitative of exponential function (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

#### KFC

I am looking a formula to compute the derivitative of $$e^{a x^2}$$ with respect to x n times, where a is a constant.

$$\frac{d^n}{dx^n}e^{a x^2}$$

I am going to find the result of above derivitative when x -> 0.

#### sutupidmath

well, take a couple of first derivatives, and see if you notice a pattern. Then if you do notice a pattern, try to prove it by induction....and after that take the limit as x-->0

P.S. I would have tried to take some derivatives myself, but it is bed time now...lol...

#### mubeenahm

The answer to this question would most probably be (2xa)^n*(e)^a(x)^2

#### Mute

Homework Helper
The answer to this question would most probably be (2xa)^n*(e)^a(x)^2
Most certainly not! You're going to have to use the chain rule every time, which is going to result in some polynomail times $e^{ax^2}$. In fact, the result is going to be related to the Hermite polynomials, defined by

$$H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$.

Getting the form desired in the OP amounts to setting $x \rightarrow iax$.

http://en.wikipedia.org/wiki/Hermite_polynomials

(Note I used the "physicist definition" of the Hermite polynomials defined on the wikipedia page).

Last edited:

#### confinement

I am looking a formula to compute the derivitative of $$e^{a x^2}$$ with respect to x n times, where a is a constant.

$$\frac{d^n}{dx^n}e^{a x^2}$$

I am going to find the result of above derivitative when x -> 0.
We define a new variable u = i sqrt(a) x:

$$\frac{d^n}{dx^n}e^{a x^2} |_{x = 0} = (i \sqrt(a))^n \frac{d^n}{du^n}e^{-u^2}|_{u = 0} = (i \sqrt(a))^n (-1)^n e^{-u^2}H_n (u) |_{u = 0} =(-2 i \sqrt(a))^n \frac{\sqrt{\pi }}{\Gamma \left(\frac{1-n}{2}\right)}$$

where the last step has used the form of the Hermite numbers:

http://mathworld.wolfram.com/HermiteNumber.html

Note also that when n is odd the Gamma function blows up and so the value we seek is zero, when n is even the imaginary unit disappears and we obtain a real result.

Last edited:

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving