[proofplan] We assume two [linear maps](/page/Linear%20Map) $\tau_1$ and $\tau_2$ both satisfy the differentiability condition at $a$. Subtracting the two representations and evaluating along rays $h = tv$ with $|v| = 1$, linearity cancels the factor of $t$, leaving the difference $(\tau_1 - \tau_2)(v)$ equal to a quantity that tends to $\mathbf{0}$ as $t \to 0^+$. Since the left side is independent of $t$, it must be $\mathbf{0}$ for every unit vector $v$, forcing $\tau_1 = \tau_2$. [/proofplan] [step:Subtract the two differentiability representations to isolate $\tau_1 - \tau_2$] Suppose $\tau_1, \tau_2 \in \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$ both satisfy the [differentiability](/page/Derivative) condition at $a$: \begin{align*} f(a + h) &= f(a) + \tau_1(h) + |h|\varepsilon_1(h), \\ f(a + h) &= f(a) + \tau_2(h) + |h|\varepsilon_2(h), \end{align*} where $\varepsilon_1, \varepsilon_2 \to \mathbf{0}$ as $h \to \mathbf{0}$. Subtracting the second equation from the first: \begin{align*} (\tau_1 - \tau_2)(h) = |h|\bigl(\varepsilon_2(h) - \varepsilon_1(h)\bigr). \end{align*} [/step] [step:Evaluate along rays $h = tv$ and use linearity to conclude $\tau_1 = \tau_2$] Fix any $v \in \mathbb{R}^m$ with $|v| = 1$. For $t > 0$ sufficiently small, $a + tv \in U$ (since $U$ is open and $a \in U$). Setting $h = tv$ in the equation above: \begin{align*} (\tau_1 - \tau_2)(tv) = t\bigl(\varepsilon_2(tv) - \varepsilon_1(tv)\bigr). \end{align*} By linearity of $\tau_1 - \tau_2$, the left side equals $t(\tau_1 - \tau_2)(v)$. Dividing both sides by $t > 0$: \begin{align*} (\tau_1 - \tau_2)(v) = \varepsilon_2(tv) - \varepsilon_1(tv). \end{align*} As $t \to 0^+$, $tv \to \mathbf{0}$, so $\varepsilon_1(tv) \to \mathbf{0}$ and $\varepsilon_2(tv) \to \mathbf{0}$. The right side tends to $\mathbf{0}$, while the left side is independent of $t$. Therefore $(\tau_1 - \tau_2)(v) = \mathbf{0}$. Since $v$ was an arbitrary unit vector, $(\tau_1 - \tau_2)(v) = \mathbf{0}$ for all $v$ with $|v| = 1$. For any $w \in \mathbb{R}^m \setminus \{\mathbf{0}\}$, write $w = |w| \cdot (w/|w|)$ and apply linearity: $(\tau_1 - \tau_2)(w) = |w| \cdot (\tau_1 - \tau_2)(w/|w|) = \mathbf{0}$. At $w = \mathbf{0}$, linearity gives $(\tau_1 - \tau_2)(\mathbf{0}) = \mathbf{0}$. Therefore $\tau_1 - \tau_2 = 0$, i.e., $\tau_1 = \tau_2$. [guided] The key idea is to use linearity to separate the dependence on direction from the dependence on magnitude. The differentiability condition involves the error $|h|\varepsilon(h)$, which is "small compared to $|h|$." The linear approximation $\tau(h)$ is exactly proportional to $|h|$ by linearity. If two linear maps $\tau_1$ and $\tau_2$ both approximate $f$ to first order, subtracting the two expansions gives $(\tau_1 - \tau_2)(h) = |h|(\varepsilon_2(h) - \varepsilon_1(h))$. The difference $(\tau_1 - \tau_2)(h)$ is "smaller than linear" in $|h|$. Evaluating at $h = tv$ with $|v| = 1$: linearity gives $(\tau_1 - \tau_2)(tv) = t(\tau_1 - \tau_2)(v)$ on the left, and $|tv|(\varepsilon_2(tv) - \varepsilon_1(tv)) = t(\varepsilon_2(tv) - \varepsilon_1(tv))$ on the right. Dividing by $t > 0$: $(\tau_1 - \tau_2)(v) = \varepsilon_2(tv) - \varepsilon_1(tv)$. The left side is a fixed vector (independent of $t$), while the right side tends to $\mathbf{0}$ as $t \to 0^+$. A constant equal to something tending to zero must be zero: $(\tau_1 - \tau_2)(v) = \mathbf{0}$. Since $v$ was an arbitrary unit vector, and any $w \neq \mathbf{0}$ can be written as $|w| \cdot (w/|w|)$, linearity gives $(\tau_1 - \tau_2)(w) = \mathbf{0}$ for all $w$. Therefore $\tau_1 = \tau_2$. This justifies writing $Df_{a}$ for "the" derivative without ambiguity. [/guided] [/step]