Partial derivative help Watch

HoldThisL
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Does this:

∂^2z / ∂x∂y
(partial squared z / partial x partial y)

Mean the differentiate with respect to y first, then with respect to x or can I differentiate either variable first?

My function is: z = 4x^2 −xy +y^2 −x^3 and I get the answer as -1 whether I differentiate x or y first but that might be a coincidence.
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I'm God
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(Original post by HoldThisL)
Does this:

∂^2z / ∂x∂y
(partial squared z / partial x partial y)

Mean the differentiate with respect to y first, then with respect to x or can I differentiate either variable first?

My function is: z = 4x^2 −xy +y^2 −x^3 and I get the answer as -1 whether I differentiate x or y first but that might be a coincidence.
99% sure that it's with respect to x first and then with respect to y.

I also got them both the same no matter how I did it when calculating Hessian, but that again may just be a coincidence.
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HoldThisL
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(Original post by I'm God)
99% sure that it's with respect to x first and then with respect to y.

I also got them both the same no matter how I did it when calculating Hessian, but that again may just be a coincidence.
That would make sense but I found a University of Surrey resource (because my university doesn't teach jack) which gave the same equation but said differentiate y first. Perhaps it is interchangeable?
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I'm God
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(Original post by HoldThisL)
That would make sense but I found a University of Surrey resource (because my university doesn't teach jack) which gave the same equation but said differentiate y first. Perhaps it is interchangeable?
Okay, I changed your equation a bit and tried it with 4x^2y - xy + y^2 -x^3y. I did it both ways and got the same answer so it probably is interchangeable, but I would still take caution.
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I'm God
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(Original post by HoldThisL)
That would make sense but I found a University of Surrey resource (because my university doesn't teach jack) which gave the same equation but said differentiate y first. Perhaps it is interchangeable?
Actually, never mind, I just found a Khan Academy video called "Symmetry of second partial derivatives" which says that the order doesn't matter. You can check it out
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HoldThisL
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(Original post by I'm God)
Actually, never mind, I just found a Khan Academy video called "Symmetry of second partial derivatives" which says that the order doesn't matter. You can check it out
Cheers
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DFranklin
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(Original post by HoldThisL)
That would make sense but I found a University of Surrey resource (because my university doesn't teach jack) which gave the same equation but said differentiate y first. Perhaps it is interchangeable?
(Original post by I'm God)
Okay, I changed your equation a bit and tried it with 4x^2y - xy + y^2 -x^3y. I did it both ways and got the same answer so it probably is interchangeable, but I would still take caution.
(Original post by I'm God)
Actually, never mind, I just found a Khan Academy video called "Symmetry of second partial derivatives" which says that the order doesn't matter. You can check it out
There's a standard result (Clairaut's theorem) that says that the order doesn't matter so long as f has continuous second partial derivatives.

This is nearly always true, but there are examples where this doesn't hold and the order does actually matter.

f(x,y) = \left\{ \begin{matrix}\dfrac{xy(x^2-y^2)}{x^2+y^2} & (x, y) \neq (0, 0) \\ 0 & (x, y) = (0, 0)\end{matrix} \right \quad

is an such an example, and you will find that at (0, 0)

\dfrac{\partial^2 f}{\partial x \partial y} = 1, while \dfrac{\partial^2 f}{\partial y \partial x} = -1

(See https://en.wikipedia.org/wiki/Symmet..._of_continuity )
Last edited by DFranklin; 1 week ago
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