# Chain Rule

romeo6
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

Cheers!

Romeo

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

romeo6
Are you really trying to find the derivative of one function with respect to the other function?

Is that not possible?

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

mjpam
Is that not possible?

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

romeo6
But if f is a function for (x,y,z) and g is a function of (x,y,z) then surely that's possible. Probably I'm wrong if I have to ask about it.

mjpam
But if f is a function for (x,y,z) and g is a function of (x,y,z) then surely that's possible. Probably I'm wrong if I have to ask about it.

Is there a way that you can justify this?

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

Amok
But if f is a function for (x,y,z) and g is a function of (x,y,z) then surely that's possible. Probably I'm wrong if I have to ask about it.

Hey, ex and x+1 are both function of x. What is the derivative of ex with respect to x+1?

romeo6
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.

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

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

Amok
Are you guys ignoring my posts on purpose :( ?

mjpam
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.

Amok
Didn't I make my point explicitly? I'm not trying to sound like a douche or anything, I actually thought the same thing as romeo6 when I first red your question.

mjpam
Didn't I make my point explicitly? I'm not trying to sound like a douche or anything, I actually thought the same thing as romeo6 when I first red your question.

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

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

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

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

Amok
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.

mjpam
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 f(x) and g(x) are both functions of x f(x) is, in some sense, dependent on g(x), because g(x)=f-1(x)). The thing that has to be kept in mind, is that, since nothing is specified about f or g, f-1 or g-1 may not themselves be functions, which means that their domains must be restricted in order for them to be differentiable.

However, with the proper assumptions made about f and/or g, the chain rule and the inverse function theorem yield:

$\frac{d(f(g^{-1}(x)))}{dx}=\frac{df}{dx}\frac{1}{\frac{dg}{dx}}$

Mentor
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
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,
$$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \text{and} \frac{\partial f}{\partial z}$$

Other notation for these partials is fx, fy, and fz.

mjpam
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.

Amok
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.

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Mentor
I think what you're talking about is a matrix of partial derivatives, which is called the Jacobian. See Jacobian. If so, that's not the same as df/dg as in the original post. The matrix is made up of the partials of each variable in the first system with respect to each variable in the second system, so we're talking about a bunch of partials of variables, with respect to other variables.

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

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

Cheers!

Romeo

Have you done Calculus III (Multivariable calculus)?

romeo6
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.

romeo6
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.

romeo6
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,
$$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \text{and} \frac{\partial f}{\partial z}$$

Other notation for these partials is fx, fy, and fz.

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.

romeo6
Have you done Calculus III (Multivariable calculus)?

Yes. I've taken some grad math also.

mjpam
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.

Amok
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.

Homework Helper
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,
$$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \text{and} \frac{\partial f}{\partial z}$$

Other notation for these partials is fx, fy, and fz.
I disagree strongly with this- you always take the derivative of a function with respect to another function! In basic Calculus , of course, that second function is the identity function, x. But asking for the derivative of f with respect to g is just asking how fast f changes relative to g. If f and g are functions of the single variable, x, then, by the chain rule
$$\frac{df}{dg}= \frac{df}{dx}\frac{dx}{dg}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}$$

If f and g are functions of the two variables x and y,
$$\frac{df}{dg}= \frac{\frac{\partial f}{\partial x}}{\frac{\partial g}{\partial x}}+ \frac{\frac{\partial f}{\partial y}}{\frac{\partial g}{\partial y}}$$