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$$ \frac{\partial \phi}{\partial \xi} ( \parallel \mathbf{x} - \xi_i \parallel) = - \frac{\partial \phi}{\partial x} ( \parallel \mathbf{x} - \xi_i \parallel) $$

from page 14 of these lecture slides, where $\phi$ is (a radial basis) function the variables $\mathbf{x}$ and $\xi_i$.

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In fact we have $\frac{\partial}{\partial x}f(x-\xi) = - \frac{\partial}{\partial \xi}f(x-\xi)$ for any function $f$, of which your formula is a special case.

This can be seen by applying the chain rule for differentiation:

$\frac{\partial}{\partial x}f(x-\xi) = f^\prime(x-\xi)\cdot\frac{\partial}{\partial x}(x-\xi) = f^\prime(x-\xi)$ whereas $\frac{\partial}{\partial \xi}f(x-\xi) = f^\prime(x-\xi)\cdot\frac{\partial}{\partial \xi}(x-\xi) = -f^\prime(x-\xi)$.

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Thanks. Just one small question before I check the answer button, what is the prime in the first and second expressions? A dummy variable which is equal to $x - \xi$? –  Olumide Jun 5 '11 at 23:23
    
It's just shorthand for the derivative of f. –  gfes Jun 5 '11 at 23:26
    
Yes, but with respect to what? Surely, it's not with respect to $x$ or $\xi$. –  Olumide Jun 6 '11 at 9:47
    
No. It's a normal derivative, not a partial one. It's $\frac{df(t)}{dt}$ evaluated at $x-\xi$. –  gfes Jun 6 '11 at 10:12
    
But if $f^\prime$ is the derivative of $f$ wrt to $t$, where $t = x - \xi$ then the final result contains $\frac{\partial f}{\partial t}$ and not $\frac{\partial f}{\partial x}$. Or is $\frac{df(t)}{dt}$ the total derivative? –  Olumide Jun 6 '11 at 11:19
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