Why is gradient co-variant ? (intuitively ) At 7:47 in this video, the professor defines a function F(x,y) on regular cartesian grid, then later defines the same function on a scaled cartesian grid (where x' = 2x and y'=2y) , now, after this he takes gradient on the function defined on the new grid. And for some reason, the gradient on the new function is four times the function in cartesian, and I can't understand why that should be the case..
I tried replicating what he did for a single variable function but my results differed quite a bit. I took the function $f(2x) = x$ then defined it on a new x' coordinate system which is squished.
$$ f(2x) = x$$
$$ 2x= x'$$
$$ f(x') = \frac{x'}{2}$$
$$ \frac{df}{dx'} = \frac{1}{2}$$
The derivative is actually half of the original function. As in the derivative scales accordingly such that when I multiply the derivative of the inner function (analogue to basis here), everything becomes the same. As in,
$$ \frac{df}{dx'} \frac{dx'}{dx} = 1$$
So, I intuitively think that the gradient in this new coordinate system should scale in such a way such that the scale up of the basis vectors is cancelled by the scaling down of the gradient. As in the gradient should be invariant of coordinating scaling and dilation.
However the regular gradient definition we all know and love is not, Why exactly does this happen? other than for the fact of it being a purely algebraic result?
Later in the video, he says that we can fix the discrepancy by doing,
$$ \nabla F = \frac{ \partial F}{\partial x} \frac{i}{|i|^2} + \frac{ \partial F}{\partial y} \frac{ j}{ |j|^2}$$
but, I can't see a systematic approach for finding that 'fix'

For a single variable note that if $f(x)=x/2$ then $df/dx=1/2 i_x$ (writing the direction of coordinates would be useful here) and after re-scaling 2 times the coordinates we have:
$$f(y)=y/2,\quad y=2x$$
and $$df/dy=1/2 i_y=1i_x$$
and this is the derivative with respect to $y$ and if we convert it to $x$ using chain rule we get:
$$df/dx=(df/dy)(dy/dx)=2\times 1i_x$$
which is 4 times of the original one i.e. $4\times 1/2i_x$.
The multi-variable case is similar.
Caution: When you write $f(2x)=x$ for an original function $f(x)=x/2$ then argue that if I take $x'=2x$ it will be $f(x')=x'/2$ is a useless cycle. In other words you are stretching coordinate first twice then stretching it in half. I.e. coming back to the original coordinates.
