How are we able to locally linearize a non-linear transform with the Jacobian matrix? After watching this video, The Jacobian Matrix, I animated what I interpreted from the video
https://imgur.com/ydSVcJI
While animating, I got the following doubt, when we look at $(1, 1)$, in the un-zoomed version, it's not fixed. I actually had to move the camera frame to be centered around $(1,1)$ in the zoomed in version. Now a linear transform has its origin (local $(x_0, y_0)$) fixed, I mean, $(1,1)$ should be mapped to itself if it were a linear transform, right?
If the local origin is not fixed, then how do we use the Jacobian matrix to approximate the non linear transform as a linear transform locally?
 A: Yes, it is true that the point itself is moved. But it is how neighbouring points are moved in relation to one another that the Jacobian records. So in the zoomed-in version, we are following $(1, 1)$ as it moves to $f(1, 1)$. We are keeping that point fixed in the middle of our screen, and watching how the points around $(1, 1)$ move during the same transformation. And it turns out that if we zoom in far enough, the points around $(1, 1)$ move in a fashion that looks almost linear.
More formally, if we have our function $f$, our point of interest $p$, and the Jacobian $J(p)$ at that point, then
$$
f(p+v)\approx f(p) + J(p)v
$$
for small vectors $v$ (where $\approx$ has a formally defined meaning). This is completely analoguous to how the regular derivative of a function works:
$$
g(x + t) \approx g(x) + g'(x) t
$$
for small numbers $t$ (or maybe you're more used to $g'(x)\approx \frac{g(x + t) - g(x)}{t}$, but that's the same thing). In fact, the derivative is simply the $1\times 1$ special case of the Jacobian.
