# Partial differentiation - chain rule

Assume I have a function f(x,y) and I transform the variables using the following:

$$\begin{equation} m = 2x + y \end{equation}$$ $$\begin{equation} n = x - y \end{equation}$$

Or any other similar linear transformation as such.

Would I be right in thinking that the following partial derivatives can be expressed in terms of m and n by the chain rule since m and n are just functions of x and y? $$\begin{equation} \frac{\partial \:f}{\partial \:x}=\frac{\partial \:m\:}{\partial \:x}\frac{\partial \:\:f\:}{\partial \:\:m}+\frac{\partial \:\:n\:}{\partial \:\:x}\frac{\partial \:\:\:f\:}{\partial \:\:\:n} \end{equation}$$ $$\begin{equation} \frac{\partial \:f}{\partial \:y}=\frac{\partial \:m\:}{\partial \:y}\frac{\partial \:\:f\:}{\partial \:\:m}+\frac{\partial \:\:n\:}{\partial \:\:y}\frac{\partial \:\:\:f\:}{\partial \:\:\:n} \end{equation}$$

I am new to multivariable calculus and I'm just curious to understand more about partial differentiation. Thank you in advance!

This was a question I had in mind after reading this website

• Your formulae are correct but I suggest reversing the order of multiplication. Of course, it doesn't matter in which order you multiply real numbers but when you have functions from R^n to R^m and you are multiplying Jacobian matrices, the order becomes significant. May 10 '20 at 23:29

In the linked pdf, notice that you have two functions, $$F$$ and $$f$$. The function $$f$$ is an expression in $$u$$ and $$v$$ that equals $$F$$ when you substitute. For example, if $$F(x,y) = x - y$$, $$u = x$$, and $$v = -y$$, you would need $$f(u,v) = u + v$$.
To maybe see how it can go wrong if you incorrectly assume $$F = f$$, let $$m = n = x$$ in your example. You would have: $$\frac{\partial f}{\partial x} = \frac{\partial x}{\partial x}\frac{\partial f}{\partial x} + \frac{\partial x}{\partial x}\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial x}.$$