Let $f: \mathbb R^{d} \to \mathbb R^{\nu}$ I am having difficulty in understanding relevance of continuity in the functional matrix $\frac{\partial(f_{1},\ldots,f_{\nu})}{\partial(x_{1},\ldots,x_{d})}$.
Why do I say this: We recently proved that for a linear map $g: \mathbb R^{d} \to \mathbb R^{\nu}, g(x):=Ax$ where $A \in \mathbb R^{d \times \nu}$ matrix, $g$ is differentiable and $(Df)(x)=A$ (*).
So surely by this statement, if the functional matrix $\frac{\partial(f_{1},\ldots,f_{\nu})}{\partial(x_{1},\ldots,x_{d})}$ exists then it has to be differentiable in $x$ and that then implies it has to be continuous in $x$?
So then why am I asked to prove that:
$\partial_{k}f_{j}$ continuous $\forall k \in \{1,\ldots,d\}$ and $\forall j \in \{1,\ldots,\nu\} \Rightarrow \frac{\partial(f_{1},\ldots,f_{\nu})}{\partial(x_{1},\ldots,x_{d})}$ continuous in $x$.
I have seen the proof for the above. But I do not understand why it cannot simply be deduced from the fact that if $\forall j \in \{1,\ldots,\nu\}$ $f_{j}$ is partially differentiable in $x$ then $\frac{\partial(f_{1},\ldots,f_{\nu})}{\partial(x_{1},\ldots,x_{d})}$ exists in $x$ and by (*) we know that it is differential in $x$ and thereby also continuous in $x$.
What is wrong with my reasoning here?