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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

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    $\begingroup$ 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. $\endgroup$ May 10, 2020 at 23:29

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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: $$ \frac{\partial f}{\partial x} = \frac{\partial x}{\partial x}\frac{\partial f}{\partial x} + \frac{\partial x}{\partial x}\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial x}. $$

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