- #1

romeo6

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e.g. df/dg=df/dx df/dy df/dz

Cheers!

Romeo

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- Thread starter romeo6
- Start date

- #1

romeo6

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e.g. df/dg=df/dx df/dy df/dz

Cheers!

Romeo

- #2

mjpam

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Are you really trying to find the derivative of one function with respect to the other function?

- #3

romeo6

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Are you really trying to find the derivative of one function with respect to the other function?

Is that not possible?

- #4

Amok

- 256

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If f is not a function g, then you derivative equals 0, if I'm not mistaken.

- #5

mjpam

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Is that not possible?

If I remember correctly, it is possible, but somewhat difficult.

- #6

romeo6

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- #7

mjpam

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Is there a way that you can justify this?

I'm just wondering if you have more than your intuition.

- #8

Amok

- 256

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Hey, e

- #9

romeo6

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Is there a way that you can justify this?

I'm just wondering if you have more than your intuition.

You're right - just intuition. Which is probably failing me.

- #10

mjpam

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You're right - just intuition. Which is probably failing me.

What is your mathematical background? Are you learning (or teaching yourself) calculus right now?

- #11

Amok

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Are you guys ignoring my posts on purpose :( ?

- #12

mjpam

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Are you guys ignoring my posts on purpose :( ?

No, but I think that if you're trying to make a point, it is best done explicitly.

- #13

Amok

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- #14

mjpam

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You answered a question with a question. (Granted that is what I did too.)

Let me think this through a little more. I didn't start with an answer in my head.

- #15

Amok

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Did you take a look at my first post in this thread?

- #16

mjpam

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Did you take a look at my first post in this thread?

I did. I just missed when I was reviewing the thread. Sorry.

- #17

Amok

- 256

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I did. I just missed when I was reviewing the thread. Sorry.

Chill, np. This threat got me thinking. Sometimes some concepts become so instinctive that thinking about them makes knots in your brain.

- #18

mjpam

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If f is not a function g, then you derivative equals 0, if I'm not mistaken.

Upon further thought, this is not true.

Since

However, with the proper assumptions made about

[itex]\frac{d(f(g^{-1}(x)))}{dx}=\frac{df}{dx}\frac{1}{\frac{dg}{dx}}[/itex]

- #19

Mark44

Mentor

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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.If I have two functions, f(x,y,z) and g(x,y,z), do I use the chain rule to calculate df/dg?

e.g. df/dg=df/dx df/dy df/dz

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

- #20

mjpam

- 79

- 0

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.

Except that is exactly what you do when you perform a variable transform.

- #21

Amok

- 256

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What do you mean by variable transform?

What Mark44 said makes sense. If you want to derive something with respect to a function you need a functional.

What Mark44 said makes sense. If you want to derive something with respect to a function you need a functional.

Last edited:

- #22

Mark44

Mentor

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- #23

mjpam

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Why does it not makes sense to define a derivative of a function with respect to another function?

- #24

chiro

Science Advisor

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e.g. df/dg=df/dx df/dy df/dz

Cheers!

Romeo

Have you done Calculus III (Multivariable calculus)?

- #25

romeo6

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If f is not a function g, then you derivative equals 0, if I'm not mistaken.

I certainly wasn't ignoring you, however I found your next post valuable, with the example of taking the derivative of x^2 wrt e^x.

- #26

romeo6

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What is your mathematical background? Are you learning (or teaching yourself) calculus right now?

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.

- #27

romeo6

- 54

- 0

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}.

This is a great answer (along with others).

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.

- #28

romeo6

- 54

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Have you done Calculus III (Multivariable calculus)?

Yes. I've taken some grad math also.

- #29

mjpam

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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.

I was just trying to gauge whether it would be appropriate to mention the set-theoretic definition of a function.

- #30

Amok

- 256

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Why does it not makes sense to define a derivative of a function with respect to another function?

I meant functional, it was a typo.

- #31

HallsofIvy

Science Advisor

Homework Helper

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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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