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.


To make sense of this, you need to understand what the word “derivative” means here.

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.


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