Partial derivative of a function defined in a rectangular area

Question

I noticed in a book that it considers a function $u(x,y)$ in the rectangular area : $x\in (0, 1)$ ,$y\in (0, 1]$, and takes there the $\partial_y$. As I am used to only to take partial derivatives in open sets, I would appreciate a clarification of the value of $\partial_y(x, 1)$. Is it $$\partial_y{u(x, 1)}=\lim_{t\to 1^{-}}\frac{u(x, t) -u(x, 1)}{t-1}$$ or something else? Thanks in advance.

Answer 1

Yes, your definition of partial derivative with respect to $y$ at $(x,1)$

$$ \partial_y{u(x, 1)}=\lim_{t\to 1^{-}}\frac{u(x, t) -u(x, 1)}{t-1}$$

is correct. Of course it is one sided because we do not have $y>1$ as part of the domain.