I am watching this video about the Jacobian matrix, and below is a picture that I am confused about.
I understand the first line $dx^i = \frac{\partial x^i}{ \partial \bar{x}^a} d\bar{x}^a$, where the Jacobian matrix allows you to convert small changes in an input space $\bar{x}^a$ to changes a position in an output space $x^i$.
When writing the 2nd line, the lecturer says "So let's take the partial derivative", with what appears to be $\frac{\partial}{\partial x^j}$.This is to show that $\frac{\partial \bar{x}^a}{ \partial x^j}$ is the inverse Jacobian matrix. (I've timestamped the video link to this point).
What I don't understand is how
$\frac{\partial}{\partial x^j} (d\bar{x}^a) = \frac{\partial \bar{x}^a}{ \partial x^j}$
For me, it seems like if you were trying to obtain the quantity $\frac{\partial \bar{x}^a}{ \partial x^j}$, then you would have to do $\frac{\partial}{\partial x^j} (\bar{x}^a)$ instead of $\frac{\partial}{\partial x^j} (\textbf{d}\bar{x}^a)$. It almost just seems like he is dividing both sides by $\partial x^j$, yet he says he is taking the partial derivative.
I've tried thinking about this operation in terms of a single-variable problem, i.e., given $y = x^3$, then $dy = 3x^2 * dx$ (for small, locally-linear values of dx), but I don't think it is valid to attempt to do
$\frac{d}{dx} (dy) = \frac{d}{dx} (3x^2 * dx)$, because isn't it invalid to take the derivative of $\textbf{d}x$ with respect to $x$? And as well, you can take the derivative of $\frac{dy}{dx}$ with respect to $x$, but I don't know if you can do $dy$ with respect to $x?
Could someone clarify what he is doing? Thanks!
I don't think the second line is supposed to be connected to the first. Just consider it a new line.
The partial derivative $\frac{\partial x}{\partial x}$ is $1$, and the partial derivative $\frac{\partial x}{\partial y}=0$ and so on, hence
$$\frac{\partial x^i}{\partial x^j}=\delta^i_j$$
But remember that
$$\bar{x}^j=\bar{x}^j(x^h)$$ with inverse $$x^h=x^h(\bar{x}^j)$$
so we can use the chain rule
$$\frac{\partial x^i}{\partial x^j}=\frac{\partial x^i}{\partial \bar{x}^l}\frac{\partial \bar{x}^l}{\partial x^j}=\delta^i_j\tag{1}$$
Of course we can do the same with $\frac{\partial \bar{x}^j }{\partial \bar{x}^l}$
$$\delta^j_l=\frac{\partial \bar{x}^j}{\partial {x}^h}\frac{\partial x^h}{\partial \bar{x}^l}\tag{2}$$
The relations $(1)$ and $(2)$ are very useful in practical caluclations.