How to Solve Second-Order Partial Derivatives

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Discussion Overview

The discussion revolves around the complexities of calculating second-order partial derivatives, particularly when a change of variables and the chain rule are involved. Participants explore the general case of functions expressed in terms of different variables and the challenges that arise in various contexts, including applications in physics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that second-order partial derivatives can be more complicated than expected and emphasizes the importance of attention to detail.
  • Another participant finds the tutorial helpful and questions the frequency of needing to use the inverse case when restating functions.
  • A different participant responds that the necessity of the inverse case depends on how the change of variables arises, providing polar coordinates as an example.
  • Another contribution highlights that in fields like Electrodynamics, the function's dependence on time and other variables complicates the situation further.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to the topic, with no consensus on the frequency or necessity of using the inverse case in different contexts.

Contextual Notes

There are unresolved aspects regarding the assumptions made about the functions involved and the specific conditions under which the inverse case is applied. The discussion also touches on the implications of time dependence in physical applications.

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Introduction
A frequent concern among students is how to carry out higher order partial derivatives where a change of variables and the chain rule are involved.  There is often uncertainty about exactly what the “rules” are.  This tutorial aims to clarify how the higher-order partial derivatives are formed in this case.
Note that in general second-order partial derivatives are more complicated than you might expect.  It’s important, therefore, to keep calm and pay attention to the details.
The General Case
Imagine we have a function ##f(u, v)## and we want to compute the partial derivatives with respect to ##x## and ##y## in terms of those with respect to ##u## and ##v##.  Here we assume that ##u, v## may be expressed as functions of ##x, y##.  The first derivative usually cause no problems.  We simply apply the chain rule:
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} \ \ \...

Continue reading...
 
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This is a really helpful reference, there's so much that can go wrong with these sorts of questions and the topic is skated over in so many resources.

How common is it to have to use the inverse case? That is, if we first have a function ##F(x,y) = f(u(x,y), v(x,y))##, is it sometimes useful to restate it in the form ##G(u,v) = g(x(u,v), y(u,v))##?
 
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etotheipi said:
How common is it to have to use the inverse case? That is, if we first have a function ##F(x,y) = f(u(x,y), v(x,y))##, is it sometimes useful to restate it in the form ##G(u,v) = g(x(u,v), y(u,v))##?
It depends how the change of variables arises. For polar coordinates we generally start from:
$$x = r\cos \phi, \ \ y = r\sin \phi$$
Which is the "inverse" case in the tutorial.

The important point is that, if you have a function of ##r, \phi##, then you must calculate the inverse functions:
$$r = \sqrt{x^2 + y^2}, \ \ \ \phi = \arctan \frac y x$$
 
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It was such a nice and clear article that I really learned from it, enjoyed and admired it. The main issue is that ##f## is not directly a function of ##x## and ##y##, and I think these cases arise so much in Electrodynamics, where fields are functions of charge/current density and position, and position in itself is a function of time, even charge/current density changes with time.
 
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