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seen Oct 8 '12 at 10:34

Oct
13
awarded  Tumbleweed
Oct
8
asked Inverse image of disjoint is disjoint?
Oct
6
comment Jacobian with Chain rules
If so, this leads to $$d(\phi\circ \psi)(x)=d\psi (\phi(x))\circ(d\phi(x))$$ If we can say that $d\psi$ is equal to $J_\psi$, and my brain is still undecided on that front, then $$d(\phi\circ \psi)(x)=J_\psi(\psi\circ h(x))\circ d(\psi\circ h(x))$$ Now, if we can expand, and again, not certain that we can: $$d(\phi\circ \psi)(x)=J_\psi(\psi\circ h(x))\circ J_\psi\circ J_h(x)$$ Someone please come along and tell me I'm wrong, this feels non-rigorous.
Oct
6
comment Jacobian with Chain rules
I don't quite follow. If we generalise, say $U\subset \mathbb{R}^n$ is open, $f:U\to \mathbb{R}^m, V\subset \mathbb{R}^m$ open and $g:V\to \mathbb{R}^k$ with $f(U)\subset V$. Letting $x\in U$, the definition of the chain rule is $$d(f\circ g)(x)=dg(f(x))\circ df(x)$$ But I still don't understand how that would fit into this notation. Are $\psi$ and $\phi$ the linear operators?!
Oct
6
asked Jacobian with Chain rules
Oct
6
awarded  Editor
Oct
6
comment Divergence Theorem to prove equality of integrals
That works, provided that I can then use the product rule for the divergence of scalar valued functions. That is, provided that u is a scalar and v is a vector field. This isn't specified, but it's also not stated, so I'm happy with this line of logic. Thanks greatly for your hint
Oct
6
awarded  Student
Oct
6
asked Divergence Theorem to prove equality of integrals