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## Main Question or Discussion Point

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:
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} \ \ \... Continue reading... • • JD_PM, Greg Bernhardt, Adesh and 3 others ## Answers and Replies etotheipi Homework Helper Gold Member 2019 Award This is a really helpful reference, there's so much that can go wrong with these sorts of questions and the topic is skated over in so many resources. How common is it to have to use the inverse case? That is, if we first have a function $F(x,y) = f(u(x,y), v(x,y))$, is it sometimes useful to restate it in the form $G(u,v) = g(x(u,v), y(u,v))$? • Greg Bernhardt Science Advisor Homework Helper Gold Member How common is it to have to use the inverse case? That is, if we first have a function $F(x,y) = f(u(x,y), v(x,y))$, is it sometimes useful to restate it in the form $G(u,v) = g(x(u,v), y(u,v))$? It depends how the change of variables arises. For polar coordinates we generally start from:$$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

• etotheipi