Keeping the beginning and end of the path fixed, we give every intermediate point an arbitrary infinitesimal displacement~$\delta x_{\sigma}$ so as to deform the path. Since \begin{align*} ds^{2} &= g_{\mu\nu}\, dx_{\mu}\, dx_{\nu}, \\ 2\, ds\, \delta(ds) &= dx_{\mu}\, dx_{\nu} \, \delta g_{\mu\nu} + g_{\mu\nu}\, dx_{\mu}\, \delta(dx_{\nu}) + g_{\mu\nu}\, dx_{\nu}\, \delta(dx_{\mu}) \\ &= dx_{\mu}\, dx_{\nu}\, \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}}\, \delta x_{\sigma} + g_{\mu\nu}\, dx_{\mu}\, d(\delta x_{\nu}) + g_{\mu\nu}\, dx_{\nu}\, d(\delta x_{\mu}). \Tag{(28.1)} \end{align*} The stationary condition is \[ \int \delta(ds) = 0, \Tag{(28.2)} \] \PageSep{60} which becomes by~\Eq{(28.1)} \[ \frac{1}{2} \int \biggl\{ \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds}\, \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}}\, \delta x_{\sigma} + g_{\mu\nu}\, \frac{dx_{\mu}}{ds}\, \frac{d}{ds}(\delta x_{\nu}) + g_{\mu\nu}\, \frac{dx_{\nu}}{ds}\, \frac{d}{ds}(\delta x_{\mu})\biggr\}\, ds = 0, \] or, changing dummy suffixes in the last two terms, \[ \frac{1}{2} \int \biggl\{ \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds}\, \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}}\, \delta x_{\sigma} + \biggl(g_{\mu\sigma}\, \frac{dx_{\mu}}{ds} + g_{\sigma\nu}\, \frac{dx_{\nu}}{ds}\biggr)\, \frac{d}{ds}(\delta x_{\sigma})\biggr\}\, ds = 0. \] Applying the usual method of partial integration, and rejecting the integrated part since $\delta x_{\sigma}$~vanishes at both limits, \[ \frac{1}{2} \int \biggl\{ \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds}\, \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}} - \frac{d}{ds} \biggl(g_{\mu\sigma}\, \frac{dx_{\mu}}{ds} + g_{\sigma\nu}\, \frac{dx_{\nu}}{ds}\biggr)\, \delta x_{\sigma}\, ds = 0. \]