Since \[ A_{\sigma} = g_{\sigma\epsilon} A^{\epsilon}, \] we have by~\Eq{(29.3)} \begin{align*} A_{\sigma\nu} &= \frac{\dd}{\dd x_{\nu}} (g_{\sigma\epsilon} A^{\epsilon}) - \{\sigma\nu, \alpha\}\, (g_{\alpha\epsilon} A^{\epsilon}) \\ &= g_{\sigma\epsilon}\, \frac{\dd A^{\epsilon}}{\dd x_{\nu}} + A^{\epsilon}\, \frac{\dd g_{\sigma\epsilon}}{\dd x_{\nu}} - [\sigma\nu, \epsilon]\, A^{\epsilon} \text{ by~\Eq{(27.4)}} \\ &= g_{\sigma\epsilon}\, \frac{\dd A^{\epsilon}}{\dd x_{\nu}} + [\epsilon\nu, \sigma]\, A^{\epsilon} \text{ by~\Eq{(27.5)}.} \end{align*} Hence multiplying through by~$g^{\mu\sigma}$, and remembering that $g^{\mu\sigma}g_{\sigma\epsilon}$~is a substitution-operator, we have \[ {A^{\mu}}_{\nu} = \frac{\dd A^{\mu}}{\dd x_{\nu}} + \{\epsilon\nu, \mu\}\, A^{\epsilon}. \Tag{(29.4)} \] \PageSep{62} This is called the covariant derivative of~$A^{\mu}$. The considerable differences \index{Covariant derivative of vector!of tensor}% \index{Derivative!covariant}% \index{Derivative!contravariant}% between the formulae \Eq{(29.3)} and \Eq{(29.4)} should be carefully noted. The tensors ${A_{\mu}}^{\nu}$ and~$A^{\mu\nu}$, obtained from \Eq{(29.3)} and~\Eq{(29.4)} by raising the second suffix, are called the \emph{contravariant derivatives} of $A_{\mu}$ and~$A^{\mu}$. We shall \index{Contravariant vectors!derivatives}% not have much occasion to refer to contravariant derivatives. \Section{30.}{Covariant derivative of a tensor} The covariant derivatives of tensors of the second rank are formed as follows--- \begin{align*} A_{\sigma}^{\mu\nu} &= \frac{\dd A^{\mu\nu}}{\dd x_{\sigma}} + \{\alpha\sigma, \mu\}\, A^{\alpha\nu} + \{\alpha\sigma, \nu\}\, A^{\mu\alpha}, \Tag{(30.1)} \\ A_{\mu\sigma}^{\nu} &= \frac{\dd A_{\mu}^{\nu}}{\dd x_{\sigma}} - \{\mu\sigma, \alpha\}\, A_{\alpha}^{\nu} + \{\alpha\sigma, \nu\}\, A_{\mu}^{\alpha}, \Tag{(30.2)} \\ A_{\mu\nu\sigma} &= \frac{\dd A_{\mu\nu}}{\dd x_{\sigma}} - \{\mu\sigma, \alpha\}\, A_{\alpha\nu} - \{\nu\sigma, \alpha\}\, A_{\mu\alpha}. \Tag{(30.3)} \end{align*} And the general rule for covariant differentiation with respect to~$x_{\sigma}$ is illustrated by the example %[** TN: Not broken in the original] \begin{multline*} A_{\lambda\mu\nu\sigma}^{\rho} = \frac{\dd}{\dd x_{\sigma}}\, A_{\lambda\mu\nu}^{\rho} - \{\lambda\sigma, \alpha\}\, A_{\alpha\mu\nu}^{\rho} - \{\mu\sigma, \alpha\}\, A_{\lambda\alpha\nu}^{\rho} \\ - \{\nu\sigma, \alpha\}\, A_{\lambda\mu\alpha}^{\rho} + \{\alpha\sigma, \rho\}\, A_{\lambda\mu\nu}^{\alpha}. \Tag{(30.4)} \end{multline*}