> [!NOTE] Lemma (Uniqueness of Derivative) > Let $U \subset \mathbb{R}^n$ be open and $f: U \rightarrow \mathbb{R}^m$. Assume $f$ is [[Fréchet Differentiation|differentiable]] at $a \in U$. Then the derivative $D f(a): \mathbb{R}^n \rightarrow \mathbb{R}^m$ is unique. ###### Proof Assume we have for $i=1,2$ $ f(a+h)=f(a)+L_i(h)+R_i(a, h) $ with $ \lim _{\substack{h \rightarrow 0 \\ h \neq 0}} \frac{\left\|R_i(a, h)\right\|}{\|h\|}=0 $ Note that $ L_1(h)-L_2(h)=R_2(a, h)-R_1(a, h) $ and thus $ \lim _{\substack{h \rightarrow 0 \\ h \rightarrow 0}} \frac{\left\|L_1(h)-L_2(h)\right\|}{\|h\|}=\lim _{\substack{h \rightarrow 0 \\ h \neq 0}} \frac{\left\|R_2(a, h)-R_1(a, h)\right\|}{\|h\|} \leq \lim _{\substack{h \rightarrow 0 \\ h \neq 0}} \frac{\left\|R_2(a, h)\right\|}{\|h\|}+\lim _{\substack{h \rightarrow 0 \\ h \neq 0}} \frac{\left\|R_1(a, h)\right\|}{\|h\|}=0 . $ Note that for any $x \in \mathbb{R}^n \backslash\{0\}$ and $t \searrow 0$ we have $t x \rightarrow 0$ as $t \searrow 0$. Thus $ 0=\lim _{t \rightarrow 0} \frac{\left\|L_1(t x)-L_2(t x)\right\|}{\|t x\|}=\frac{\left\|L_1(x)-L_2(x)\right\|}{\|x\|} $ which yields $L_1(x)=L_2(x)$. Since this holds for any $x \in \mathbb{R}^n \backslash\{0\}$ we have $L_1=L_2$.