Ronan
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# 9 Actions

 Oct13 awarded Tumbleweed Oct8 asked Inverse image of disjoint is disjoint? Oct6 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. Oct6 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?! Oct6 asked Jacobian with Chain rules Oct6 awarded Editor Oct6 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 Oct6 awarded Student Oct6 asked Divergence Theorem to prove equality of integrals