1. ## Partial Derivatives

Hello Forumites,

The partial derivatives of order r of an analytic function $f(x_1,...x_n)$ of n variables do not depend on the order of differentiation but only on the number of times that each variables appears.And hence there exists $\binom{n+r-1}{r}$ different partial derivatives of rth order. A function of three variables has fifteen derivatives of fourth order and 21 derivatives of fifth order.
What does this means? If any one knows, he may reply.

2. ## Re: Partial Derivatives

There are some "she's" on this forum too … just sayin'

3. ## Re: Partial Derivatives

Originally Posted by Debsta
There are some "she's" on this forum too … just sayin'
Hello,
If any forumite knows the answer, post it into this thread. Suppose $f(x,y,z)=3x^4+3y^2-3z^2-56$. Now in this function there are three variables,how many partial derivatives can be formed?

4. ## Re: Partial Derivatives

"A function of three variables has fifteen derivatives of fourth order."

Lets say your variables are x, y and z.

For the fourth order partial derivatives, you need to look at what you can take these derivatives with respect to. Order makes no difference.

So you could take the fourth derivative wrt x then x then x then x again - for simplicity (and so I don't have to write in LaTex) let's call that xxxx.

So basically you are finding how many strings of 4 you can make using the variables x, y and/or z.

xxxx
xxxy
xxyy
xyyy

yyyy
yyyz
yyzz
yzzz

zzzz
zzzx
zzxx
zxxx

xyzz
xyyz
xxyz .... 15 of them! n+r-1 = 3+4-1=6 ; r=4 ; 6C4 =15

5. ## Re: Partial Derivatives

I feel like it's important to point out that mixed partial derivatives are not necessarily equal. Here is an example: https://en.wikipedia.org/wiki/Symmet..._of_continuity.

Since you said the function is analytic, though, it is infinitely differentiable. And the polynomial you give is infinitely differentiable. Because of this, Debsta's answer is correct.

The number of ways to choose r things from a group of n (with replacement, order does not matter) is the formula you gave: $\displaystyle \binom{n+r-1}{r}$.

- Hollywood