1. To obtain the covariant derivative of any tensor~$A_{....}^{....}$ with respect to~$x_{\sigma}$, \index{Derivative!covariant}% we take first the ordinary derivative \[ \frac{\dd}{\dd x_{\sigma}} A_{....}^{....}\,; \] and for \emph{each} covariant suffix~$A_{..\mu.}^{....}$, we add a term \[ - \{\mu\sigma, \alpha\} A_{..\alpha.}^{....}; \] and for \emph{each} contravariant suffix~$A_{....}^{..\mu.}$, we add a term \[ + \{\alpha\sigma, \mu\} A_{....}^{..\mu.}. \] 2. The covariant derivative of a product is formed by covariant differentiation of each factor in turn, by the same rule as in ordinary differentiation. 3. The fundamental tensor~$g_{\mu\nu}$ or~$g^{\mu\nu}$ behaves as though it were a constant in covariant differentiation. 4. The covariant derivative of an invariant is its ordinary derivative. 5. In taking second, third or higher derivatives, the order of differentiation is not interchangeable\footnotemark.\footnotetext {This is inserted here for completeness; it is discussed later.} \Section{31.}{Alternative discussion of the covariant derivative} By~\Eq{(23.22)} \[ g_{\mu\nu}' = \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \frac{\dd x_{\beta}}{\dd x_{\nu}'}\, g_{\alpha\beta}. \] Hence differentiating \[ \frac{\dd g_{\mu\nu}'}{\dd x_{\lambda}'} = g_{\alpha\beta} \left\{ \frac{\dd^{2} x_{\alpha}}{\dd x_{\lambda}'\, \dd x_{\mu}'}\, \frac{\dd x_{\beta}}{\dd x_{\nu}'} + \frac{\dd^{2} x_{\alpha}}{\dd x_{\lambda}'\, \dd x_{\nu}'}\, \frac{\dd x_{\beta}}{\dd x_{\mu}'}\right\} + \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \frac{\dd x_{\beta}}{\dd x_{\nu}'}\, \frac{\dd x_{\gamma}}{\dd x_{\lambda}'}\, \frac{\dd g_{\alpha\beta}}{\dd x_{\gamma}}. \Tag{(31.11)} \]