[proofplan] We prove both directions of the equivalence. The forward direction applies the [Chain Rule](/theorems/323) to $f_j = \pi_j \circ f$, using the fact that each coordinate projection $\pi_j$ is linear and hence [differentiable](/page/Derivative). The reverse direction assembles the componentwise error terms into a single vector-valued error and shows its norm vanishes using the fact that $|\varepsilon(h)|^2 = \sum_j \varepsilon_j(h)^2$. [/proofplan] [step:Deduce component differentiability from differentiability of $f$ via the chain rule] Suppose $f$ is [differentiable](/page/Derivative) at $a$ with derivative $\tau \in \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$. Each coordinate projection $\pi_j: \mathbb{R}^n \to \mathbb{R}$ defined by $\pi_j(y_1, \ldots, y_n) = y_j$ is linear, hence differentiable everywhere with $D(\pi_j)_y = \pi_j$ for all $y$. By the [Chain Rule](/theorems/323) applied to $f_j = \pi_j \circ f$, each $f_j$ is differentiable at $a$ with \begin{align*} D(f_j)_a = \pi_j \circ \tau. \end{align*} [/step] [step:Reconstruct differentiability of $f$ from differentiability of each component] Suppose each $f_j$ is [differentiable](/page/Derivative) at $a$ with derivative $\tau_j \in \mathcal{L}(\mathbb{R}^m, \mathbb{R})$, so that \begin{align*} f_j(a + h) = f_j(a) + \tau_j(h) + |h|\varepsilon_j(h), \end{align*} with $\varepsilon_j(h) \to 0$ as $h \to \mathbf{0}$. Define $\tau \in \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$ by $\tau(h) = (\tau_1(h), \ldots, \tau_n(h))$ and set $\varepsilon(h) = (\varepsilon_1(h), \ldots, \varepsilon_n(h))$. Then \begin{align*} f(a + h) = f(a) + \tau(h) + |h|\varepsilon(h). \end{align*} Since $|\varepsilon(h)|^2 = \sum_{j=1}^n \varepsilon_j(h)^2 \to 0$ as $h \to \mathbf{0}$ (each summand tends to $0$), we have $\varepsilon(h) \to \mathbf{0}$. Therefore $f$ is differentiable at $a$ with $Df_a = \tau$. [guided] Why can we assemble the componentwise expansions so cleanly? The key point is that the Euclidean norm on $\mathbb{R}^n$ satisfies $|\varepsilon|^2 = \sum_j \varepsilon_j^2$, so the vector error $\varepsilon(h)$ tends to $\mathbf{0}$ if and only if each scalar component $\varepsilon_j(h)$ tends to $0$. This is where we use the product structure of $\mathbb{R}^n$. More explicitly, define $\tau \in \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$ by $\tau(h) = (\tau_1(h), \ldots, \tau_n(h))$. This is indeed linear because each $\tau_j$ is linear. Setting $\varepsilon(h) = (\varepsilon_1(h), \ldots, \varepsilon_n(h))$, the componentwise differentiability expansions combine to \begin{align*} f(a + h) = f(a) + \tau(h) + |h|\varepsilon(h). \end{align*} We verify the error condition: $|\varepsilon(h)|^2 = \sum_{j=1}^n \varepsilon_j(h)^2$. Since each $\varepsilon_j(h) \to 0$, the sum of squares tends to $0$, so $\varepsilon(h) \to \mathbf{0}$. This confirms $f$ is [differentiable](/page/Derivative) at $a$ with $Df_a = \tau$, and $Df_a(h) = (D(f_1)_a(h), \ldots, D(f_n)_a(h))$. [/guided] [/step]