My analysis textbook - Marsden Elementary Analysis pg. 155 has this definition:
Theorem 1: Let $A$ be an open set in $\mathbb{R}^n$ and suppose $f : A \mapsto \mathbb{R}^m$ is differentiable at $x_0$. Then $Df(x_0)$ is uniquely determined by $f.$
Not sure what it means and how it can be applied. Any help would be appreciated.
"The" derivative at $x_0$ is, a priori, a linear map $T$ satisfying $$f(x_0+h)=f(x_0)+T\cdot h+\epsilon(h),$$ with $\epsilon(h)$ such that $\Vert\epsilon(h)\Vert/\Vert h\Vert \to 0$ as $h \to 0$.
What the text means by "uniquely determined" is that, if such $T$ exists, it is unique. One way to see this is to consider an arbitrary $h \in \mathbb{R}^n$, and evaluate the expression at $th$, where $t$ is a real number. If $t$ is small enough, the expression is meaningful since $A$ is open. We then have $$f(x_0+th)=f(x_0)+tT\cdot h+\epsilon(th).$$ Dividing by $t$ and handling this a bit, we get $$\frac{f(x_0+th)-f(x_0)}{t}=T \cdot h +\frac{\epsilon(th)}{t}.$$ Letting $t \to 0$, we have $$T \cdot h=\lim\limits_{t \to 0}\frac{f(x_0+th)-f(x_0)}{t}.$$ Therefore, the value of $T \cdot h$, if $T$ exists, is determined by $f$ alone. Since this holds for every $h$, we have that $T$ is uniquely determined by $f$, and we can drop the quotation marks at the "The" on the beginning. Depending on what terminologies you already know, you will recognize that the expression above is a kind of directional derivative (taking $h=e_i$ gives the usual partial derivatives, and this is also a proof of the fact that the derivative will take on the Jacobian matrix form if we see it as a matrix w.r.t. the canonical basis).