# Symmetry in second order partial derivatives and chain rule

Tags:
1. Mar 15, 2015

### powerof

When can I do the following where $h_{i}$ is a function of $(x_{1},...,x_{n})$?

$\frac{\partial}{\partial x_{k}}\frac{\partial f(h_{1},...,h_{n})}{\partial h_{m}}\overset{?}{=}\frac{\partial}{\partial h_{m}}\frac{\partial f(h_{1},...,h_{n})}{\partial x_{m}}\overset{\underbrace{chain\ rule}}{=}\frac{\partial}{\partial h_{m}} \sum_{l=1}^{n} \frac{\partial f(h_{1},...,h_{n})}{\partial h_{l}}\frac{\partial h_{l}}{\partial x_{k}}$

And why? Thank you for your time.

2. Mar 15, 2015

### andrewkirk

I'm pretty sure this won't work.

I don't think the middle bit is well-defined. Partial derivatives operate on a functional form, not on a value. In this problem $f$ is given as a function of the $h_i$, so the partial derivatives $\frac{\partial f(h_{1},...,h_{n})}{\partial h_i}$ are well-defined. But if you want to take partial derivatives of $f$ with respect to an $x_k$ you need to change the functional form of $f$ and it will become a new function $g$. The most natural interpretation is to replace each $h_i$ by its expression in terms of the $x_j$s. But when you've done that you no longer have any $h_i$s in the functional form, which is $\frac{\partial g(x_{1},...,x_{n})}{\partial x_k}$, so $\frac{\partial}{\partial h_i}$ will be zero.

In some cases you might be able to turn the $x_j$s back into $h_i$s, and then do the, $\frac{\partial}{\partial h_i}$, but to do that you'd need there to be a bijection between $x_1,...,x_n$ and $h_1,...,h_n$ and in general there won't be.

Even if there is a bijection, the sought equality may not hold or be meaningless.

eg consider the case $f=h_1h_2$ where $h_1=x_1+x_2, h_2=x_1-x_2$. Then the LHS of your equation is $\frac{\partial^2 f}{\partial x_1 \partial h_1}=1$. To calculate the middle bit, we first express $f$ in terms of the $x$s as $f(h_1,h_2)=g(x_1,x_2)=(x_1+x_2)(x_1-x_2)=x_1^2-x_2^2$. Then $\frac{\partial}{\partial x_1}$ is $2x_1$. How are we now going to take a partial derivative of this with respect to $h_1$? We could write $2x_1=2(h_1 -y)$, in which case the answer is 2, or we could write $2x_1=2(h_2+y)$ in which case the answer is zero. So there is not a unique answer.