Let $E$ and $F$ be Banach spaces and let $U$ be an open subset of $E$. Suppose $g:U \to F $. $g$ is continuous at $x_0$ if there exists a linear transformation, $T_{x_0}$, such that $$ \lim \limits_{x_0 \to 0}\dfrac{||g(x_0+h)-g(x_0)-T_{x_0}(h)||}{||h||}. $$ Then $Dg_{x_0}=T_{x_0}$ is then the Frechet derivative of $g$ at $x_0$.
I have a question about this with two examples from this video (https://www.youtube.com/watch?v=RKHx1vQdZko).
For the first example we compute the Frechet derivative of $f(x)=x^2$ and we get that $T_x(h) = h(2x)$. For the second example we find the Frechet derivative of $f(x,y)=(xy , x+y)$ and here we get that $T_{(x,y)}(h,v)=\begin{bmatrix} y & x \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} h \\ v \\ \end{bmatrix}$ .
But the definition says that $Df_{x_0}=T_{x_0}$ so does that mean that $Df_{x_0}=T_{x_0}(1)$? It seems to be the case for both of these examples. I believe this is always but am unsure. The reason why I think this is that the transformation is linear so $T_{x_0}(h)=h(Df_{x_0})$ since $T$ linearly maps $h$ so if we let $h=1$ then we are just left with $Df_{x_0}$. Does my reasoning make sense?