By~\Eq{(23.12)} \[ A_{\mu}' = \frac{\dd x_{\epsilon}}{\dd x_{\mu}'}\, A_{\epsilon}. \Tag{(31.4)} \] Hence differentiating \begin{align*} \frac{\dd A_{\mu}'}{\dd x_{\nu}'} &= \frac{\dd^{2} x_{\epsilon}}{\dd x_{\mu}'\, \dd x_{\nu}'}\, A_{\epsilon} + \frac{\dd x_{\epsilon}}{\dd x_{\mu}'}\, \frac{\dd x_{\delta}}{\dd x_{\nu}'}\, \frac{\dd A_{\epsilon}}{\dd x_{\delta}} \\ &= \left(\{\mu\nu, \rho\}'\, \frac{\dd x_{\epsilon}}{\dd x_{\rho}'} - \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \frac{\dd x_{\beta}}{\dd x_{\nu}'}\, [\alpha\beta, \epsilon]\right) A_{\epsilon} + \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \frac{\dd x_{\beta}}{\dd x_{\nu}'}\, \frac{\dd A_{\alpha}}{\dd x_{\beta}} \Tag{(31.5)} \end{align*} by \Eq{(31.3)} and changing the dummy suffixes in the last term. Also by~\Eq{(23.12)} \[ A_{\epsilon}\, \frac{\dd x_{\epsilon}}{\dd x_{\rho}'} = A_{\rho}'. \] Hence \Eq{(31.5)}~becomes \[ \frac{\dd A_{\mu}'}{\dd x_{\nu}'} - \{\mu\nu, \rho\}' A_{\rho}' = \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \frac{\dd x_{\beta}}{\dd x_{\nu}'} \left(\frac{\dd A_{\alpha}}{\dd x_{\beta}} - \{\alpha\beta, \epsilon\}\, A_{\epsilon}\right)\!, \Tag{(31.6)} \] showing that \[ \frac{\dd A_{\mu}}{\dd x_{\nu}} - \{\mu\nu, \rho\}\, A_{\rho} \] obeys the law of transformation of a covariant tensor. We thus reach the result~\Eq{(29.3)} by an alternative method. A tensor of the second or higher rank may be taken instead of~$A_{\mu}$ in~\Eq{(31.4)}, and its covariant derivative will be found by the same method. \Section{32.}{Surface-elements and Stokes's theorem} {\Loosen Consider the outer product~$\Sigma^{\mu\nu}$ of two different displacements $dx_{\mu}$ and $\delta x_{\nu}$. The tensor $\Sigma^{\mu\nu}$ will be unsymmetrical in $\mu$ and~$\nu$. We can decompose any such tensor into the sum of a symmetrical part $\frac{1}{2}(\Sigma^{\mu\nu} + \Sigma^{\nu\mu})$ and an antisymmetrical part $\frac{1}{2}(\Sigma^{\mu\nu} - \Sigma^{\nu\mu})$.}