Derivative of a $x^T$ with respect to $x$

Question

I am trying to understand $\frac{d x^T}{d x}$ where $x$ is a $n\times 1$ vector. I have read and tried to understand this answer already but I have some other questions. As far as I understand, when we generalize it, the derivation of a function (either a scalar or vector function) with respect to a column vector can be considered as:

$$ \frac{df(x)}{dx} = \begin{bmatrix}\frac{df(x)}{dx_1}\\...\\ \frac{df(x)}{dx_n}\end{bmatrix}$$

Then, If we consider the function to be the function that gives us the transpose of the input vector, then we should have:

$$\frac{df(x)}{dx} = \begin{bmatrix}\frac{dx^T}{dx_1}\\...\\ \frac{dx^T}{dx_n}\end{bmatrix}$$

Then this means we need to calculate:

$$\frac{df(x)}{dx} = \begin{bmatrix}\frac{\begin{bmatrix}x_1 \ ... \ x_n \end{bmatrix}}{dx_1}\\...\\ \frac{\begin{bmatrix}x_1 \ ... \ x_n \end{bmatrix}}{dx_n}\end{bmatrix}$$

My intution says this means we need to convert this to:

$$\frac{df(x)}{dx} = \begin{bmatrix} \frac{x_1}{dx_1} \ \frac{x_2}{dx_1} \ ... \ \frac{x_n}{dx_1} \\ ... \\ \frac{x_1}{dx_n} \ \frac{x_2}{dx_n} \ ... \ \frac{x_n}{dx_n} \end{bmatrix}$$

which gives us the identity matrix. Is this intuition true? If so how does this work and why? I don't quite understand what's going on theoritically when I take the derivative of a row vector w.r.t a column vector or a row vector w.r.t. a scalar.