Understanding the chain rule for differentiation operators

Question

Suppose I want to transform a partial derivative operator from spherical to Cartesian coordinates. I have found the following relation based on the chain rule here:

$$ \frac{\partial }{\partial \theta } = \frac{\partial x}{\partial \theta} \frac{\partial}{\partial x} + \frac{\partial y}{\partial \theta} \frac{\partial}{\partial t} + \dots $$

As I know from calculus, the chain rule is commonly defined, for example when want to take derivative of some 'functions' wrt to some variables.

So, how can the chain rule be defined for the derivative operators? I don't exactly understand the chain rule in this context for the system transformation.

Thank you.

Answer 1

Suppose we have $f = f(r,s)$ where $r=r(x,y)$ and $s=s(x,y)$. Then, by the usual chain rule for functions you mentioned: $$\frac{\partial}{\partial x}f(r(x,y),s(x,y)) = \frac{\partial r}{\partial x}\frac{\partial f(r,s)}{\partial r} + \frac{\partial s}{\partial x}\frac{\partial f(r,s)}{\partial x}$$

Thus, if $\frac{\partial}{\partial x}$ is viewed as an operator acting on $f$, you can see $\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial s}{\partial x}\frac{\partial}{\partial x}$ also as an operator acting on $f$. This is what we mean by: $$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} + \frac{\partial s}{\partial x}\frac{\partial}{\partial x}$$ which can be interpreted as a chain rule for operators.

Answer 2

Transforming spherical coordinates to Cartesian coordinates is just a mapping:

\begin{align} x(r, \theta, \varphi) & = r \sin \theta \cos \varphi \\ y(r, \theta, \varphi) & = r \sin \theta \sin \varphi \\ z(r, \theta, \varphi) & = r \cos \theta. \end{align}

Given $f\equiv g\circ(x,y,z)$, it follows from the chain rule (assuming the appropriate differentiability) that $$ \frac{\partial f}{\partial\theta}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial\theta}+\frac{\partial g}{\partial z}\frac{\partial z}{\partial\theta}. $$