Is there an explicit relationship between the eigenvalues of a matrix and its derivative?

It is not true in general that the eigenvalues of $B(x)$ are the derivatives of those of $A(x)$. And this even for some square matrices of dimension $2 \times 2$.

Consider the matrix

$$A(x) = \begin{pmatrix} 1& -x^2\\ -x &1 \end{pmatrix}$$ It’s characteristic polynomial is $\chi_{A(x)}(t)=t^2-2t+1-x^3$, which has for roots $1\pm x ^{3/2}$ for $x>0$. Those are the eigenvalues of $A(x)$.

The derivative of $A(x)$ is $$B(x) = \begin{pmatrix} 0& -2x\\ -1 &0 \end{pmatrix}$$ and it’s characteristic polynomial is $\chi_{B(x)}(t)=t^2-2x$, whose roots are $\pm \sqrt{2} x^{1/2}$for $x>0$.

We get a counterexample as the derivative of $1+x^{3/2}$ is not $\sqrt{2}x^{1/2}$.

However in the special case of upper triangular matrices (that you consider in your original question) the eigenvalues of the matrix derivative are indeed the derivatives of the eigenvalues.

It can be shown that the eigenvalues of the derivative of the matrix cannot be derived from the eigenvalues of the original matrix. Example: $$ A_1 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \;\;\; , \;\;\; A_2 = \begin{pmatrix} 0 & e^x \\ e^{-x} & 0 \end{pmatrix} $$ Both of the matrices above have the eigenvalues $-1$ and $1.$ However, the derivative of the first matrix has the eigenvalue $0$ (with multiplicity 2), while the derivative of the second matrix has the eigenvalues $i$ and $-i.$

Just given the eigenvalues $-1$ and $1$, there is no way of telling which matrix they originate from, hence no way of getting the eigenvalues of the derivative.