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I've taken plenty of calculus (believe it or not), it's been a few years now though, and I've not used it for a while.What is your mathematical background? Are you learning (or teaching yourself) calculus right now?

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I've taken plenty of calculus (believe it or not), it's been a few years now though, and I've not used it for a while.What is your mathematical background? Are you learning (or teaching yourself) calculus right now?

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This is a great answer (along with others).This really doesn't make much sense. You don't calculate the derivative of a function with respect to some other function, but you do calculate the derivative of a function with respect to one of its variables. Here g is a function, not a variable, so df/dg is nonsensical.

For another thing, both functions here have multiple variables, so instead of df/dx, df/dy, and df/dz, you would be working with partial derivatives,

[tex]\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \text{and} \frac{\partial f}{\partial z}[/tex]

Other notation for these partials is f_{x}, f_{y}, and f_{z}.

Actually, I've started thinking about it in terms of orthogonal basis. Would you're base vectors not have to be orthogonal to take a derivative with respect to something else? If you started taking derivatives wrt a basis that was some function...well, that would be a bit of a nightmare.

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Yes. I've taken some grad math also.Have you done Calculus III (Multivariable calculus)?

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I was just trying to gauge whether it would be appropriate to mention the set-theoretic definition of a function.I've taken plenty of calculus (believe it or not), it's been a few years now though, and I've not used it for a while.

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I meant functional, it was a typo.Why does it not makes sense to define a derivative of a function with respect to another function?

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I disagree strongly with this- youThis really doesn't make much sense. You don't calculate the derivative of a function with respect to some other function, but you do calculate the derivative of a function with respect to one of its variables. Here g is a function, not a variable, so df/dg is nonsensical.

For another thing, both functions here have multiple variables, so instead of df/dx, df/dy, and df/dz, you would be working with partial derivatives,

[tex]\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \text{and} \frac{\partial f}{\partial z}[/tex]

Other notation for these partials is f_{x}, f_{y}, and f_{z}.

[tex]\frac{df}{dg}= \frac{df}{dx}\frac{dx}{dg}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}[/tex]

If f and g are functions of the two variables x and y,

[tex]\frac{df}{dg}= \frac{\frac{\partial f}{\partial x}}{\frac{\partial g}{\partial x}}+ \frac{\frac{\partial f}{\partial y}}{\frac{\partial g}{\partial y}}[/tex]

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