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Symmetry in second order partial derivatives and chain rule

  1. Mar 15, 2015 #1
    When can I do the following where ##h_{i}## is a function of ##(x_{1},...,x_{n})##?

    [itex]\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}}[/itex]

    And why? Thank you for your time.
     
  2. jcsd
  3. Mar 15, 2015 #2

    andrewkirk

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    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.
     
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