Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In ${\mathbb R}^n$, let $F$ be a smooth one-to-one mapping of $\Omega$ onto some open set $\Omega'$, where $\Omega\subset{\mathbb R}^n$ is open. Set $y=F(x)$. Assume that the Jacobian matrix $J_x=[(\partial y_i/\partial x_j)(x)]$ is nonsingular for $x\in\Omega$. We have $$\frac{\partial}{\partial x_j}=\sum\frac{\partial y_i}{\partial x_j}\frac{\partial}{\partial y_i}.$$

Here are my questions:

Is there a neat way to calculate $$\frac{\partial^2}{\partial x_j\partial x_k}?$$ After several steps trial, I am completely confused. More generally, what is $\partial_x^{\alpha}$ in terms of the $y$ coordinate system? Here $$\partial_x^{\alpha}:=\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n}}.$$

share|cite|improve this question
Unless there are simplifying assumptions about the mapping in question, the only way to calculate mixed partials is with the repeated chain rule. That makes the last question like asking what the general chain rule $D^k (f\circ g)$ is for arbitrary $k$, but then you go and split even that inevitable monstrosity into $n$ different $f$'s and $g$'s. You're not going to get anything human-manageable from that angle. – anon Aug 6 '11 at 3:45
One more reason not to think in coordinates. – Qiaochu Yuan Aug 6 '11 at 14:40
Like anon says, the "difficulty" is already present in the one-dimensional case. You should think about that before you deal with change of coordinates in higher dimensions. – Willie Wong Aug 12 '11 at 2:32
up vote 2 down vote accepted

It's a straightforward computation, you just have to do it: $$\frac{\partial^2f}{\partial x_j\partial x_k}=\frac{\partial}{\partial x_j}\left(\sum_i\frac{\partial y_i}{\partial x_k}\frac{\partial f}{\partial y_i}\right)=\sum_i\frac{\partial^2 y_i}{\partial x_j\partial x_k}\frac{\partial f}{\partial y_i}+\sum_i\frac{\partial y_i}{\partial x_k}\frac{\partial}{\partial x_j}\left(\frac{\partial f}{\partial y_i}\right)$$

Now all that is left is to also replace $\frac{\partial}{\partial x_j}$ so you get

$$\frac{\partial^2}{\partial x_j\partial x_k}=\sum_i\frac{\partial^2 y_i}{\partial x_j\partial x_k}\frac{\partial}{\partial y_i}+\sum_{h,i}\frac{\partial y_i}{\partial x_k}\frac{\partial y_h}{\partial x_j}\frac{\partial^2}{\partial y_h\partial y_i}$$ Notice that $\frac{\partial}{\partial x_j}$ is a derivation, so it can be defined by a (coordinate independent) vector field.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.