[proofplan] We substitute the [differentiability](/page/Derivative) expansion $f(a + tu) = f(a) + tDf_a(u) + |t||u|\varepsilon(tu)$ and divide by $t$ to compute the directional derivative, showing it equals $Df_a(u)$. Then linearity of $Df_a$ expresses the total derivative in terms of partial derivatives via the standard basis decomposition. [/proofplan] [step:Compute the directional derivative from the differentiability expansion] By [differentiability](/page/Derivative) of $f$ at $a$, for any $u \in \mathbb{R}^m \setminus \{\mathbf{0}\}$ and $t \neq 0$ sufficiently small: \begin{align*} f(a + tu) = f(a) + Df_a(tu) + |tu|\varepsilon(tu) = f(a) + tDf_a(u) + |t||u|\varepsilon(tu), \end{align*} where linearity of $Df_a$ gives $Df_a(tu) = tDf_a(u)$, and $\varepsilon(tu) \to \mathbf{0}$ as $t \to 0$. Rearranging and dividing by $t$: \begin{align*} \frac{f(a + tu) - f(a)}{t} = Df_a(u) + \frac{|t|}{t}|u|\varepsilon(tu). \end{align*} The factor $|t|/t = \pm 1$ is bounded and $\varepsilon(tu) \to \mathbf{0}$, so the second term vanishes as $t \to 0$. Therefore the [limit](/page/Limit) exists and $D_{u}f(a) = Df_a(u)$. [/step] [step:Express the total derivative in terms of partial derivatives via basis decomposition] For any $h = \sum_{i=1}^m h_i e_i$, linearity of $Df_a$ gives: \begin{align*} Df_a(h) = Df_a\left(\sum_{i=1}^m h_i e_i\right) = \sum_{i=1}^m h_i Df_a(e_i) = \sum_{i=1}^m h_i D_if(a), \end{align*} where the last equality uses the result of the previous step with $u = e_i$: $Df_a(e_i) = D_{e_i}f(a) = D_if(a)$. [/step]