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

Cheers!

Romeo

- Thread starter romeo6
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- #1

- 54

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

Cheers!

Romeo

- #2

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

- #3

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Is that not possible?Are you really trying to find the derivative of one function with respect to the other function?

- #4

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

- #5

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If I remember correctly, it is possible, but somewhat difficult.Is that not possible?

- #6

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- #7

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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

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Hey, e

- #9

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You're right - just intuition. Which is probably failing me.Is there a way that you can justify this?

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

- #10

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What is your mathematical background? Are you learning (or teaching yourself) calculus right now?You're right - just intuition. Which is probably failing me.

- #11

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Are you guys ignoring my posts on purpose :( ?

- #12

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No, but I think that if you're trying to make a point, it is best done explicitly.Are you guys ignoring my posts on purpose :( ?

- #13

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- #14

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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

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

- #16

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I did. I just missed when I was reviewing the thread. Sorry.Did you take a look at my first post in this thread?

- #17

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Chill, np. This threat got me thinking. Sometimes some concepts become so instinctive that thinking about them makes knots in your brain.I did. I just missed when I was reviewing the thread. Sorry.

- #18

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Upon further thought, this is not true.If f is not a function g, then you derivative equals 0, if I'm not mistaken.

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

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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

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

- #21

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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

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- #23

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

- #24

chiro

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Have you done Calculus III (Multivariable calculus)?

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

Cheers!

Romeo

- #25

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

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