# Questions about Partial Differentiation Operations

Ahmed Mehedi
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) ?

Gold Member
1) yes, it's the chain rule
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.

• Ahmed Mehedi
Ahmed Mehedi
1) yes, it's the chain rule
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.
Thanks a lot for your kind clarification!

• dRic2
Homework Helper
Gold Member
2022 Award
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
$$\frac{\partial C}{\partial r} = 2$$

• dRic2, Ahmed Mehedi and etotheipi
Ahmed Mehedi
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
$$\frac{\partial C}{\partial r} = 2$$
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: $$\frac{\partial I}{\partial C}=\frac{1}{\frac{\partial C}{\partial I}}$$

Ahmed Mehedi
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
$$\frac{\partial C}{\partial r} = 2$$
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.

Homework Helper
Gold Member
2022 Award
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: $$\frac{\partial I}{\partial C}=\frac{1}{\frac{\partial C}{\partial I}}$$
That's not true for partial derivatives. Take polar coordinates:
$$x = r \cos \theta, \ \ r = \sqrt{x^2 + y^2}$$
This gives:
$$\frac{\partial x}{\partial r} = \cos \theta \ \ \text{and} \ \ \frac{\partial r}{\partial x} = \frac x r = \cos \theta$$

• Ahmed Mehedi
Ahmed Mehedi
That's not true for partial derivatives. Take polar coordinates:
$$x = r \cos \theta, \ \ r = \sqrt{x^2 + y^2}$$
This gives:
$$\frac{\partial x}{\partial r} = \cos \theta \ \ \text{and} \ \ \frac{\partial r}{\partial x} = \frac x r = \cos \theta$$

• PeroK and Ahmed Mehedi
Ahmed Mehedi
• etotheipi and PeroK
Gold Member
@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
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

• PeroK
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2022 Award
@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
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.

• etotheipi
The relation $$\frac{\partial y}{\partial x} = \frac{1}{\frac{\partial x}{\partial y}}$$ is generally not true, except in the case that the same variables are being held constant in each differentiation. So $$\left(\frac{\partial x}{\partial r}\right)_{\theta} = \frac{1}{\left(\frac{\partial r}{\partial x}\right)_{\theta}}$$ but as has been shown above, $$\left(\frac{\partial x}{\partial r}\right)_{\theta} \neq \frac{1}{\left(\frac{\partial r}{\partial x}\right)_{y}}$$

• Ahmed Mehedi and PeroK
Gold Member
Well i though that It was implicitly assumed that you have to keep the same variables constant.

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

• Ahmed Mehedi
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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• Ahmed Mehedi
Gold Member
Yes, but that Is not what I meant. In the first partial derivative you are keeping the volume constant, while in the second the pressure.

An example of what I was describing is this:
$$T = \left(\frac {\partial E}{\partial S} \right)_V$$
And
$$\frac 1 T = \left(\frac {\partial S}{\partial E} \right)_V$$

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

• Ahmed Mehedi and dRic2
Gold Member
Yes, that is what I was saying. I thought it was an obvious thing to assume

• etotheipi