Can you show the derivative X-transpose.X wrt x?

Certainly! Let's assume that $X$ is a column vector (a matrix with only one column). In that case, $X^T$ is the transpose of $X$, and $X \cdot X^T$ results in a square matrix.
Let $X = [x_1, x_2, \ldots, x_n]^T$, where $x_i$ represents the elements of the column vector $X$.
Now, let's find the derivative of $X^T \cdot X$ with respect to $X$. The result will be a matrix of the same size as $X$.
$\frac{\partial (X^T \cdot X)}{\partial X}$
$\frac{\partial (X^T \cdot X)}{\partial X} = \frac{\partial}{\partial X} \left( \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \cdot [x_1, x_2, \ldots, x_n] \right)$
$= \frac{\partial}{\partial X} \left( x_1^2 + x_2^2 + \ldots + x_n^2 \right)$
$= \begin{bmatrix} 2x_1 \\ 2x_2 \\ \vdots \\ 2x_n \end{bmatrix}$
So, the derivative of $X^T \cdot X$ with respect to $X$ is a column vector with each element being twice the corresponding element of $X$:
$\frac{\partial (X^T \cdot X)}{\partial X} = \begin{bmatrix} 2x_1 \\ 2x_2 \\ \vdots \\ 2x_n \end{bmatrix}$

If you want, you can take the 2 out from the column vector in the final result.