Given a real symmetric matrix $A$ with entries depending on $t$, the derivative $p$-th eigenvalue with respect to $t$ is given by
$$ \lambda_p' = v_p^T A'v_p $$
where $A'$ denotes the derivative of matrix $A$. This can be derived by premultiplying $$ A' v_p + A v_p' = \lambda_p' v_p + \lambda_p v_p' $$
with $v_p^T$, imposing that the eigenvectors have unit length (and thus $v_p \cdot v_p' = 0$), and making use of the fact that $A$ is symmetric.
Say I want to use the same approach on the eigendecomposition of $A$ $$ A V = V \Lambda \\ A'V + A V' = V' \Lambda + V \Lambda' \\ V^T A'V + V^T A V' = V^T V' \Lambda + V^T V \Lambda' \\ V^T A'V + V^T A V' = V^T V' \Lambda + \Lambda' \\ \Lambda' = V^T A'V + V^T A V' - V^T V' \Lambda \\ \Lambda' = V^T A'V + \Lambda V^T V' - V^T V' \Lambda \\ $$
also
$$ I' = (V^T V)' = {V^T}'V + V^T V' = 0 $$
However, I don't see how I could use this statement to simplify the earlier expression. I feel like I'm missing something really obvious.
How can I express the derivative of the matrix of eigenvalues $\Lambda$ in terms of the derivative of $A$?