Proof that the derivative is unique? Given a subset $\Omega$ of $\mathbb{R}^n$ and a function $\sigma: \Omega \to \mathbb{R}^n$, we define its derivative at $x$ to be a linear operator $\sigma'(x)$ such that $$  \lim_{y \to 0} \frac{||\,\sigma(x + y) - \sigma(x) - \sigma'(x)y\,||}{||\,y\,||} = 0 $$
If this operator exists (i.e. if $\sigma$ is differentiable at $x$) then how do we prove that $\sigma'(x)$ is unique?
 A: Suppose there are linear operators $\sigma_{1}'$ and $\sigma_{2}'$ such that
$$
   \lim_{\|y\|\rightarrow 0}\frac{\|\sigma(x+y)-\sigma(x)-\sigma_{j}'(x)y\|}{\|y\|} =0, \\\;\;\; j =1,2.
$$
Because $\sigma_{2}'(x)$ and $\sigma_{1}'(x)$ are linear,
$$
   (\sigma_{2}'(x)-\sigma_{1}'(x))(y)= \\
= \{\sigma(x+y)-\sigma(x)-\sigma_{1}'(x)y\}\\
-\{\sigma(x+y)-\sigma(x)-\sigma_{2}'(x)y\}
$$
Applying the triangle inequality gives
$$
     \lim_{\|y\|\rightarrow 0}\frac{\|(\sigma_{1}'-\sigma_{2}')(y)\|}{\|y\|}=0
$$
In particular, for any non-zero vector $y$, one has $\alpha y \rightarrow 0$ as the scalar $\alpha$ tends to 0. Therefore, by linearity of $\sigma_{1}'-\sigma_{2}'$,
$$
      0= \lim_{\alpha\rightarrow 0}\frac{\|(\sigma_{1}'-\sigma_{2}')(\alpha y)\|}{\|\alpha y\|} \\= \lim_{\alpha\rightarrow 0}\frac{\|(\sigma_{1}'-\sigma_{2}')(y)\|}{\|y\|}
   \\= \frac{\|(\sigma_{1}'-\sigma_{2}')(y)\|}{\|y\|}.
$$
So $\|(\sigma_{2}'-\sigma_{1}')(y)\|=0$ for all non-zero vectors $y$, which implies that $\sigma_{2}'=\sigma_{1}'$.
