Consider a bivariate, real-valued function $u \colon X \times Y \to \mathbb{R}$, with $X, Y \subset \mathbb{R}$. Assume that $u$ is differentiable in the second variable on $X \times Y$, hence $\partial_y u(x,y)$ exists for all $(x,y) \in X \times Y$. Note that I do not want to make any additional hypothesis on the existence of $\partial_x u$.
Define an operator $$ U \colon Y \to V, \qquad y \mapsto u(\cdot,y), $$ where $V$ is a Banach space of functions defined on $X$.
This question is about finding a connection, if at all possible, between the partial derivative $\partial_y u$ of $u$ and the Fréchet derivative $d U$of $U$. For instance, assume that one proves that the following operator is bounded on $Y$ to $V$, $$ A \colon y \mapsto \partial_y u(\cdot,y), $$ then one may hope that the Fréchet derivative of $U$ at $y \in Y$ coincides with $A(y)$, that is, $dU(y) = A(y)$.
To me, however, this may require some additional conditions. From the definition of partial derivative of $u$ we know that $$ u(x,y+h) - u(x,y) - \partial_y u(x,y)h = o(|h|) \qquad \text{as $h \to 0$ for all $x \in X$} $$ hence $$ \lim_{|h| \to 0} \frac{|u(x,y+h) - u(x,y) - \partial_y u(x,y)h|}{|h|} = 0 \qquad \text{for all $x \in X$,} $$ and what we must deduce is $$ \lim_{|h| \to 0} \frac{\|u(\cdot,y+h) - u(\cdot,y) - \partial_y u(\cdot,y)h\|_{V}}{|h|} = 0. $$
I can't see why this would be true without further assumptions, for instance, for $V = C(X)$. Do you have any ideas when this is true, or should I just give up and think on a case-by-case basis?
