# covariant derivatives: of contravariant vector from covariant derivative covariant vector

Having proved eqn for covariant derivative covariant vector, want to get eqn for Covariant derivative contravariant vector,using metric tensor $${T^\delta } = {g^{\delta \alpha }}{T_\alpha }$$ $$\implies {T^\delta }_{;\beta } = {({g^{\delta \alpha }}{T_\alpha })_{;\beta }}$$ $$= ({g^{\delta \alpha }}_{;\beta }){T_\alpha } + {g^{\delta \alpha }}({T_{\alpha ;\beta }})$$ First term drops out as metric covariantly constant $$= {g^{\delta \alpha }}({T_{\alpha ,\beta }} - {\Gamma ^\mu }_{\alpha \beta }{T_\mu })$$ $$= {T^\delta }_{,\beta } - {g^{\delta \alpha }}{\Gamma ^\mu }_{\alpha \beta }{T_\mu }$$ $$= {T^\delta }_{,\beta } - {g^{\delta \alpha }}{\Gamma ^\delta }_{\alpha \beta }{T_\delta }$$ $$= {T^\delta }_{,\beta } - {\Gamma ^\delta }_{\alpha \beta }{T^\alpha }$$ Obviously I’ve lost a sign in the connection term, but don’t see how

You've actually got two mistakes in there. First, from the 4th line to the 5th $${T^{\delta}}_{ ,\beta}=(g^{\delta \alpha}T_{\alpha})_{,\beta}\neq g^{\delta\alpha}T_{\alpha,\beta}$$ And second, in the 6th line you've got three $$\delta$$ indices in one term, which is not allowed.
Here's an alternative way to do it. Take any $$V_{\alpha}$$, then $$V_{\alpha}T^{\alpha}$$ is a scalar, so acting on it with the covariant derivative is just partial differentiation. Expanding both the partial and the covariant derivative with the chain rule gives $$V_{\alpha,\beta}T^{\alpha}+V_{\alpha}{T^{\alpha}}_{,\beta}=V_{\alpha;\beta}T^{\alpha}+V_{\alpha}{T^{\alpha}}_{;\beta}$$ Now use the formula for $$V_{\alpha;\beta}$$ that you already now, cancel a term, and use the fact that the result must be valid for all $$V_{\alpha}$$.