- #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) ?
Thanks a lot for your kind clarification!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.
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) ?
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: $$\frac{\partial I}{\partial C}=\frac{1}{\frac{\partial C}{\partial I}}$$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.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
$$\frac{\partial C}{\partial r} = 2$$
That's not true for partial derivatives. Take polar coordinates: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: $$\frac{\partial I}{\partial C}=\frac{1}{\frac{\partial C}{\partial I}}$$
Thanks a lot! Your answers are very concrete and very helpful!PeroK said:That's not true for partial derivatives. Take polar coordinates:
$$x = r \cos \theta, \ \ r = \sqrt{x^2 + y^2}$$
This gives:
$$\frac{\partial x}{\partial r} = \cos \theta \ \ \text{and} \ \ \frac{\partial r}{\partial x} = \frac x r = \cos \theta$$
Ahmed Mehedi said: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?
Thanks a lot for your suggestions! Those notes are really well-written and very helpful!etotheipi said: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.
PeroK said:That's not true for partial derivatives. Take polar coordinates:
$$x = r \cos \theta, \ \ r = \sqrt{x^2 + y^2}$$
This gives:
$$\frac{\partial x}{\partial r} = \cos \theta \ \ \text{and} \ \ \frac{\partial r}{\partial x} = \frac x r = \cos \theta$$
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.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
dRic2 said:Well i though that It was implicitly assumed that you have to keep the same variables constant.
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.
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.
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.
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.
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.