Questions about Partial Differentiation Operations

In summary: If ##r = \frac x {\cos \theta}## and I differentiate wrt to x keeping ##\theta## constant the relation holds....
  • #1
Ahmed Mehedi
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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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  • #2
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.
 
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  • #3
dRic2 said:
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!
 
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  • #4
Ahmed Mehedi said:
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
$$\frac{\partial C}{\partial r} = 2$$
 
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  • #5
PeroK said:
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: $$\frac{\partial I}{\partial C}=\frac{1}{\frac{\partial C}{\partial I}}$$
 
  • #6
PeroK said:
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.
 
  • #7
Ahmed Mehedi said:
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:
$$\frac{\partial x}{\partial r} = \cos \theta \ \ \text{and} \ \ \frac{\partial r}{\partial x} = \frac x r = \cos \theta$$
 
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  • #8
PeroK said:
That's not true for partial derivatives. Take polar coordinates:
$$x = r \cos \theta, \ \ r = \sqrt{x^2 + y^2}$$
This gives:
$$\frac{\partial x}{\partial r} = \cos \theta \ \ \text{and} \ \ \frac{\partial r}{\partial x} = \frac x r = \cos \theta$$
Thanks a lot! Your answers are very concrete and very helpful!
 
  • #11
@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
PeroK said:
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
 
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  • #12
dRic2 said:
@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 thisif ##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.
 
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  • #13
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, $$\left(\frac{\partial x}{\partial r}\right)_{\theta} \neq \frac{1}{\left(\frac{\partial r}{\partial x}\right)_{y}}$$
 
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  • #14
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.
 
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  • #15
dRic2 said:
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
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
$$\frac 1 T = \left(\frac {\partial S}{\partial E} \right)_V$$
 
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  • #17
Yes that's okay, since you're holding ##V## constant both times.
 
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  • #18
Yes, that is what I was saying. I thought it was an obvious thing to assume
 
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Related to Questions about Partial Differentiation Operations

1. What is partial differentiation?

Partial differentiation is a mathematical operation used to find the rate of change of a multivariable function with respect to one of its variables, while holding all other variables constant.

2. Why is partial differentiation important?

Partial differentiation is important because it allows us to analyze how a function changes with respect to specific variables, which is crucial in many fields of science such as physics, economics, and engineering.

3. How is partial differentiation different from ordinary differentiation?

Ordinary differentiation deals with functions of a single variable, while partial differentiation deals with functions of multiple variables. In partial differentiation, we only consider the change in the function with respect to one variable, while holding all others constant.

4. What is the purpose of the partial derivative symbol (∂)?

The partial derivative symbol (∂) is used to represent the operation of partial differentiation. It indicates that we are taking the derivative with respect to a specific variable, while holding all other variables constant.

5. Can partial differentiation be applied to any type of function?

Yes, partial differentiation can be applied to any type of function, as long as it has multiple variables. It is a fundamental tool in calculus and is used in various applications in science and engineering.

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