[proofplan] We start from the [Characterisation of Twice Differentiability](/theorems/330), which provides the bilinear expansion $Df_{a + h}(k) = Df_a(k) + f''(a)(h, k) + |h|\varepsilon(h, k)$. We set $k = e_j$ to express $D_jf(a + h)$ in terms of $f''(a)$, then set $h = te_i$ and take the difference quotient $t \to 0$ to recover $D_i D_j f(a) = f''(a)(e_i, e_j)$. [/proofplan] [step:Express $D_jf$ in terms of the total derivative] By the [Directional Derivatives from the Total Derivative](/theorems/326) theorem, for every $x$ where $f$ is [differentiable](/page/Derivative): \begin{align*} D_jf(x) = Df_x(e_j). \end{align*} [/step] [step:Substitute into the bilinear expansion and take the difference quotient in $e_i$] By the [Characterisation of Twice Differentiability](/theorems/330), setting $k = e_j$ and $h = te_i$: \begin{align*} Df_{a + te_i}(e_j) = Df_a(e_j) + t \, f''(a)(e_i, e_j) + |t| \, \varepsilon(te_i, e_j), \end{align*} where we used $|te_i| = |t|$ and the linearity of $f''(a)$ in the first argument. By the previous step, this reads: \begin{align*} D_jf(a + te_i) = D_jf(a) + t \, f''(a)(e_i, e_j) + |t| \, \varepsilon(te_i, e_j). \end{align*} Dividing by $t \neq 0$ and taking $t \to 0$: \begin{align*} \frac{D_jf(a + te_i) - D_jf(a)}{t} = f''(a)(e_i, e_j) + \frac{|t|}{t}\varepsilon(te_i, e_j). \end{align*} Since $|t|/t = \pm 1$ is bounded and $\varepsilon(te_i, e_j) \to \mathbf{0}$ as $t \to 0$, the [limit](/page/Limit) of the left side exists and equals $f''(a)(e_i, e_j)$. By definition, the left side is $D_i(D_jf)(a) = D_i D_j f(a)$. [/step]