I'm having trouble understanding the proof/solution below (please see photo, I also wrote out the problem below). I highlighted the part of my problem in red (in the picture attached). Basically I'm not sure what identity they use to get the Kronecker delta after differentiating or whether they use something else to get that part.(adsbygoogle = window.adsbygoogle || []).push({});

"Since A^{/}_{[itex]\nu[/itex]}is covariant vector, it satisfies

A^{/}_{[itex]\nu[/itex]}= ([itex]\partial[/itex]x^{[itex]\rho[/itex]}[itex]/[/itex][itex]\partial[/itex]x^{/}^{[itex]\nu[/itex]})A_{[itex]\rho[/itex]}

A^{/}_{[itex]\rho[/itex]}= ([itex]\partial[/itex]x^{/}^{[itex]\mu[/itex]}[itex]/[/itex][itex]\partial[/itex]x^{[itex]\rho[/itex]})A^{/}_{[itex]\mu[/itex]}

This next part gives me trouble with the relation:

([itex]\partial[/itex]A_{[itex]\rho[/itex]}[itex]/[/itex][itex]\partial[/itex]A^{/}_{[itex]\nu[/itex]}) = ([itex]\partial[/itex]x^{/}_{[itex]\nu[/itex]}[itex]/[/itex][itex]\partial[/itex]x_{[itex]\rho[/itex]}) [itex]\delta[/itex]^{[itex]\nu[/itex]}_{[itex]\mu[/itex]}

I'm lost as to how the kronecker delta appears. I know that

([itex]\partial[/itex]x^{[itex]\nu[/itex]}[itex]/[/itex][itex]\partial[/itex]x_{[itex]\mu[/itex]}) = [itex]\delta[/itex]^{[itex]\nu[/itex]}_{[itex]\mu[/itex]}

[Mod Note: Over-sized image removed. See attachment. Hoot]

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# Covariant vector differentiation problem with kronecker delta?

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