Suppose I have a function $f(x,y)$ and wish to find $\frac{\partial^2 f}{\partial x \partial y}$.

However, it is easier for me to do so if I use the substitutions $u = g(x)$ and $v=h(y)$

My question is: What is the rule for using substitution to find mixed partial derivatives. I wasn't able to find any looking on Google. Is it okay to just use:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial u \partial v} \frac{\partial u}{\partial x} \frac{\partial v}{\partial y}$ ?

The derivatives I am trying to find are very messy (asymmetric copulas for various distributions), and involve using substitutions like $u = -logx$ etc.

  • $\begingroup$ In fact your function $f(x,y)$ can be rewritten in terms of $u$ and $v$ as it folows: $$f(x,y) = f(g^{-1}(x),h^{-1}(y))$$ and then you can derive it with respect $u$ and $v$. For the sake of clarity $a=h^{-1}(b)$ represents the inverse of the function $h(a)=b$ $\endgroup$ – HBR Dec 15 '16 at 11:01
  • $\begingroup$ If I do that it's just as messy to deal with, plus I still need the final derivative to be with respect to $x$ and $y$. So I'm unsure how that helps. $\endgroup$ – Patty Dec 15 '16 at 11:04
  • $\begingroup$ Sorry for the typo in my last comment. It should have been : $$f(x,y) = f(g^{-1}(u), h^{-1}(v))$$ What I proposed, definitely does not help. It is obvious that the function you have $f(x,y)$ is a function of $x$ and $y$, what you cannot do is define $g(x) = u$ and $h(y) = v$ and say that $f(x,y) = f(u,v)$ $\endgroup$ – HBR Dec 15 '16 at 11:09
  • $\begingroup$ I think I guess it. Let me suppose your function is easily written as $f(u,v)$ with $u=u(x)$ and $v=v(y)$, then you can do what you propose: $$\partial_x\partial_yf(u(x),v(y)) = \partial_u\partial_vf(u,v)d_xud_yv$$ $\endgroup$ – HBR Dec 15 '16 at 11:17

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