- #1

Ahmed Mehedi

- 39

- 5

- TL;DR Summary
- Partial Differentiation

1) If we have two functions C(y, r) and I(y, r) can we write: ∂C/∂I×∂I/∂r=∂C/∂r ? Can we also write ∂I/∂C=1/(∂C/∂I) ?

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- Thread starter Ahmed Mehedi
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- #1

Ahmed Mehedi

- 39

- 5

- TL;DR Summary
- Partial Differentiation

1) If we have two functions C(y, r) and I(y, r) can we write: ∂C/∂I×∂I/∂r=∂C/∂r ? Can we also write ∂I/∂C=1/(∂C/∂I) ?

- #2

dRic2

Gold Member

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2) In practice you often have to deal with "good" functions and it works. If you want to be super-rigorous I think you can find some examples where it is not true, but you usually do not run into such problems in physics or engineering.

- #3

Ahmed Mehedi

- 39

- 5

Thanks a lot for your kind clarification!

2) In practice you often have to deal with "good" functions and it works. If you want to be super-rigorous I think you can find some examples where it is not true, but you usually do not run into such problems in physics or engineering.

- #4

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- 15,688

Summary::Partial Differentiation

1) If we have two functions C(y, r) and I(y, r) can we write: ∂C/∂I×∂I/∂r=∂C/∂r ? Can we also write ∂I/∂C=1/(∂C/∂I) ?

This is not right. The chain rule for partial derivatives requires that you have the functions defined in terms of the other variables. In this case, you must imagine that you have ##C## as a function of ##I## and ##r## and then apply the multi-variable chain rule. We can take an example:

$$C(y, r) = y + r \ \ \text{and} \ \ I(y, r) = y - r \ \ \text{then} \ \ C(I, r) = I + 2r$$

Then:

$$\frac{\partial C}{\partial I}\frac{\partial I}{\partial r} = (1)(-1) = -1$$

And

$$\frac{\partial C}{\partial r} = 2$$

- #5

Ahmed Mehedi

- 39

- 5

Thanks a lot brother for your nice and concrete explanation and thanks for your time. Now, I get some insight about the problem I am facing. I have another question: In the aforementioned context is it legit to write: $$\frac{\partial I}{\partial C}=\frac{1}{\frac{\partial C}{\partial I}}$$This is not right. The chain rule for partial derivatives requires that you have the functions defined in terms of the other variables. In this case, you must imagine that you have ##C## as a function of ##I## and ##r## and then apply the multi-variable chain rule. We can take an example:

$$C(y, r) = y + r \ \ \text{and} \ \ I(y, r) = y - r \ \ \text{then} \ \ C(I, r) = I + 2r$$

Then:

$$\frac{\partial C}{\partial I}\frac{\partial I}{\partial r} = (1)(-1) = -1$$

And

$$\frac{\partial C}{\partial r} = 2$$

- #6

Ahmed Mehedi

- 39

- 5

So, in this case $$\frac{\partial C}{\partial I}\frac{\partial I}{\partial r} \neq \frac{\partial C}{\partial r}$$ Then how does the multi-variable chain rule look like in this case? Again thanking you for your precious time.This is not right. The chain rule for partial derivatives requires that you have the functions defined in terms of the other variables. In this case, you must imagine that you have ##C## as a function of ##I## and ##r## and then apply the multi-variable chain rule. We can take an example:

$$C(y, r) = y + r \ \ \text{and} \ \ I(y, r) = y - r \ \ \text{then} \ \ C(I, r) = I + 2r$$

Then:

$$\frac{\partial C}{\partial I}\frac{\partial I}{\partial r} = (1)(-1) = -1$$

And

$$\frac{\partial C}{\partial r} = 2$$

- #7

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That's not true for partial derivatives. Take polar coordinates:Thanks a lot brother for your nice and concrete explanation and thanks for your time. Now, I get some insight about the problem I am facing. I have another question: In the aforementioned context is it legit to write: $$\frac{\partial I}{\partial C}=\frac{1}{\frac{\partial C}{\partial I}}$$

$$x = r \cos \theta, \ \ r = \sqrt{x^2 + y^2}$$

This gives:

$$\frac{\partial x}{\partial r} = \cos \theta \ \ \text{and} \ \ \frac{\partial r}{\partial x} = \frac x r = \cos \theta$$

- #8

Ahmed Mehedi

- 39

- 5

Thanks a lot! Your answers are very concrete and very helpful!That's not true for partial derivatives. Take polar coordinates:

$$x = r \cos \theta, \ \ r = \sqrt{x^2 + y^2}$$

This gives:

$$\frac{\partial x}{\partial r} = \cos \theta \ \ \text{and} \ \ \frac{\partial r}{\partial x} = \frac x r = \cos \theta$$

- #9

So, in this case $$\frac{\partial C}{\partial I}\frac{\partial I}{\partial r} \neq \frac{\partial C}{\partial r}$$ Then how does the multi-variable chain rule look like in this case?

Might I also suggest:

https://www.physicsforums.com/insights/demystifying-chain-rule-calculus/

https://www.physicsforums.com/insights/how-to-solve-second-order-partial-derivatives/

I think you have a lot of similar concerns that I had.

- #10

Ahmed Mehedi

- 39

- 5

Thanks a lot for your suggestions! Those notes are really well-written and very helpful!Might I also suggest:

https://www.physicsforums.com/insights/demystifying-chain-rule-calculus/

https://www.physicsforums.com/insights/how-to-solve-second-order-partial-derivatives/

I think you have a lot of similar concerns that I had.

- #11

dRic2

Gold Member

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That's not true for partial derivatives. Take polar coordinates:

$$x = r \cos \theta, \ \ r = \sqrt{x^2 + y^2}$$

This gives:

$$\frac{\partial x}{\partial r} = \cos \theta \ \ \text{and} \ \ \frac{\partial r}{\partial x} = \frac x r = \cos \theta$$

if ##r = \frac x {\cos \theta}## and I differentiate wrt to x keeping ##\theta## constant the relation holds. What you did is differentiating wrt to ##x## but keeping only ##y## constant. And that is a different matter

- #12

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That's a special case where you have only three variables: ##x, r, \theta##. If we fix ##\theta##, then we have effectively a single-variable function relating ##r## and ##x##. In general, you can't simply invert partial derivatives.@PeroK Thanks for correcting my wrong statements (for the first question I naively assumed that C was expressed as function of I which could not happen, as you pointed out), but I don't understand this

if ##r = \frac x {\cos \theta}## and I differentiate wrt to x keeping ##\theta## constant the relation holds. What you did is differentiating wrt to ##x## but keeping only ##y## constant. And that is a different matter

- #13

- #14

dRic2

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In thermodynamics you do this trick all the time when you deal with thermodynamic potentials and their derivatives.

- #15

Well i though that It was implicitly assumed that you have to keep the same variables constant.

I don't have much experience so I could be mistaken, but usually it is important in thermodynamics to keep track of which variables you hold constant. You can hold different variables constant without any issues. For a non-ideal gas, for instance, $$C_{V} = \left( \frac{\partial U}{\partial T} \right)_{V} \neq \left( \frac{\partial U}{\partial T} \right)_{P}$$ I would have thought it be generally incorrect to state $$\left(\frac{\partial U}{\partial T}\right)_{V} = \frac{1}{\left(\frac{\partial T}{\partial U}\right)_{P}}$$for exactly the same reasons given above.

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- #16

dRic2

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An example of what I was describing is this:

$$T = \left(\frac {\partial E}{\partial S} \right)_V$$

And

$$\frac 1 T = \left(\frac {\partial S}{\partial E} \right)_V$$

- #17

Yes that's okay, since you're holding ##V## constant both times.

- #18

dRic2

Gold Member

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- 225

Yes, that is what I was saying. I thought it was an obvious thing to assume

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