How to compute the variation of two covariant derivatives?

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SUMMARY

The discussion focuses on computing the variation of two covariant derivatives in modified gravity models, specifically involving the perturbation of field equations. The participant presents their derivation of the expression $$\delta(\nabla_\rho \nabla_\nu [F'(G) R_{\mu}^{\hphantom{\mu} \rho}])$$ and seeks validation of their reasoning. The consensus confirms the correctness of their approach while highlighting the importance of applying the product rule for derivatives, which states that $$\delta(\nabla_\rho \nabla_\nu F(G)) = \nabla_\rho \delta(\nabla_\nu F(G)) + \nabla_\nu \delta(\nabla_\rho F(G))$$ to simplify the expression.

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balaustrada
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Homework Statement
I need to compute the variation of two covariant derivatives applied to a tensor, and I'm not sure how it should be computed.
Relevant Equations
$$ \delta ( \nabla_\rho \nabla_\nu \left[ F'(G) R_{\mu}^{\hphantom{\mu} \rho} \right])$$
I'm working with modfied gravity models and I need to consider the perturbation of field equations. I have problems with the term were I have two covariant derivatives, I'm not sure if I'm doing it right.

I have:
$$\delta(\nabla_\rho \nabla_\nu \left[F'(G)R_{\mu}^{\hphantom{\mu} \rho}\right])$$
where F(G) a function of the Gauss-Bonet comination.

What I consider, is the following:
$$ \delta (\nabla_\rho V^\mu) = \delta ( \partial_\rho V^\mu + \Gamma^\mu_{\rho \xi} V^\xi) = \partial_\rho \delta(V^\mu) + \Gamma^\mu_{\rho \xi} \delta(V^\xi) + \delta (\Gamma^{\mu}_{\rho \xi}) V^\xi \\
= \nabla_\rho \delta(V^{\mu}) + \delta(\Gamma^\mu_{\rho \xi} V^\xi) $$

Applied with two covariant derivatives gives:
$$\delta(\nabla_\rho \nabla_\nu \left[ F'(G) R_{\mu}^{\hphantom{\mu} \rho} \right]) = \nabla_\rho \delta(\nabla_\nu \left[ F'(G) R_{\mu}^{\hphantom{\mu} \rho} \right]) - \delta(\Gamma^\xi_{\rho \nu}) \left(\nabla_\xi \left[ F'(G) R_{\mu}^{\hphantom{\mu} \rho} \right] \right) - \delta(\Gamma^\xi_{\rho \mu})\left(\nabla_\nu \left[ F'(G) R_{\xi}^{\hphantom{\mu} \rho} \right] \right)
+ \delta(\Gamma^\rho_{\nu \xi})\left(\nabla_\nu \left[ F'(G) R_{\mu}^{\hphantom{\mu} \xi} \right] \right)$$

and finally:

$$ \delta(\nabla_\rho \nabla_\nu \left[F'(G)R_{\mu}^{\hphantom{\mu} \rho}\right]) = \nabla_\rho \left[ \nabla_\nu \delta( \left[ F'(G) R_{\mu}^{\hphantom{\mu} \rho} \right] ) - \delta(\Gamma^\xi_{\nu \mu}) \left[F'(G)R_{\xi}^{\hphantom{\xi} \rho} \right] + \delta(\Gamma^\rho_{\nu \xi}) \left[F'(G) R_{\mu}^{\hphantom{\xi} \xi} \right] \right] \\
- \delta(\Gamma^\xi_{\rho \nu}) \left(\nabla_\xi \left[ F'(G) R_{\mu}^{\hphantom{\mu} \rho} \right] \right) - \delta(\Gamma^\xi_{\rho \mu})\left(\nabla_\nu \left[ F'(G) R_{\xi}^{\hphantom{\mu} \rho} \right] \right) + \delta(\Gamma^\rho_{\nu \xi})\left(\nabla_\nu \left[ F'(G) R_{\mu}^{\hphantom{\mu} \xi} \right] \right)$$

Is this reasoning true?
 
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Am I missing something? The answer to your question is yes, your reasoning is correct. The only thing you may be missing is the product rule for derivatives, which states that $$\delta(\nabla_\rho \nabla_\nu F(G)) = \nabla_\rho \delta(\nabla_\nu F(G)) + \nabla_\nu \delta (\nabla_\rho F(G))$$This rule can be used to simplify some of the terms in your expression.
 

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