# Why is continuity of partials necessary for Jacobian to be the derivative?

Jut having the partials is not enough, but partials being continuous is enough to guarantee that the Jacobian is the derivative.

My question is: "Why is it necessary for the partial derivatives to be continuous? Isn't just the existence of partials sufficient?"

According to what I know, the Jacobian is a just a matrix consisting of the partial derivatives.

A function $f \colon \mathbf{R}^n \to \mathbf{R}^m$ is said to be differentiable at the point $a \in \mathbf{R}^n$ if there is a linear transformation $L \colon \mathbf{R}^n \to \mathbf{R}^m$ such that $$f(a+h) - f(a) = L(h) + R(h)$$ with a remainder term $R(h)$ which tends to zero faster than $|h|$ as $h \to 0$, i.e., $$\lim_{h \to 0} \frac{R(h)}{|h|} = \lim_{h \to 0} \frac{f(a+h) - f(a) - L(h)}{|h|} = 0 . \tag{*}$$ If this is satisfied, the linear transformation $L$ is called the derivative of $f$ at $a$. (Sometimes it's called the total derivative, to distinguish it from the partial derivatives of $f$.)
One can think of the linear transformation $L$ as just a matrix of size $n \times m$, and in fact it turns out that if the derivative exists at all, then this matrix has to be the Jacobian matrix (the $n \times m$ matrix of partial derivatives at the point $a$).
But merely the fact that the Jacobian exists is not sufficient to guarantee that the derivative exists. There are counterexamples, like $f(x,y)=17$ if $x=0$ or $y=0$, and $f(x,y)=43$ otherwise; the partials of $f$ at the origin both exist (they are zero), but $f$ has no (total) derivative there, and in fact $f$ doesn't even fulfill the weaker condition of being continuous at the origin.
However, there's a theorem that says that if the partials exist and are continuous (in some open set containing $a$), then the derivative exists. And this is what your teacher was referring to.