Directional derivative and steepest ascent along the gradient

Define the single-variable function $g(t) = f(a + t\mathbf{u})$. By definition of the directional derivative, $D_{\mathbf{u}}f(a) = g'(0)$. Since $f$ is totally differentiable at $a$, we have $f(a + t\mathbf{u}) - f(a) = \nabla f(a) \cdot (t\mathbf{u}) + o(|t|)$ as $t \to 0$. Dividing by $t$ and taking the limit $t \to 0$ yields $D_{\mathbf{u}}f(a) = \nabla f(a) \cdot \mathbf{u}$.

By the geometric formula for the Euclidean inner product and the unit-length condition $\|\mathbf{u}\| = 1$, we obtain $\nabla f(a) \cdot \mathbf{u} = \|\nabla f(a)\|\,\|\mathbf{u}\|\cos\theta = \|\nabla f(a)\|\cos\theta$. Since $-1 \le \cos\theta \le 1$, the maximum value $\|\nabla f(a)\|$ is achieved uniquely at $\theta = 0$, where $\mathbf{u} = \frac{\nabla f(a)}{\|\nabla f(a)\|}$, and the minimum $-\|\nabla f(a)\|$ is achieved at $\theta = \pi$.