Currently, I am facing the following problem.
Let $t \mapsto H (t)$ be a continuously differentiable, symmetric matrix-valued function. I would like to calculate the following derivative. Is there any general method to calculate it?$$\frac{{\rm d}}{{\rm d} t} \lambda_{\max}(H(t))$$
$ \def\l{\lambda} \def\qiq{\quad\implies\quad} \def\qif{\quad\iff\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} $Consider the eigenvector $u$ associated with $\l=\l_{max}\,$ of the matrix $\,H=H^T$
Start with the eigenvalue equation $$\eqalign{ Hu = \l u \qif \c{u^TH=\l u^T} \\ }$$ Calculate the differential, multiply by $u^T$ and isolate $d\l$ $$\eqalign{ dH\:u + H\:du &= \l\:du + u\:d\l \\ u^TdH\:u + \c{u^TH}\,du &= \l\,{u^Tdu} + u^Tu\:d\l \\ u^TdH\:u + (\c{\l u^T}\,du) &= ({\l u^Tdu}) + u^Tu\:d\l \\ }$$ $$\eqalign{ u^TdH\:u &= u^Tu\:d\l \qiq \boxed{ \;d\l = \frac{u^TdH\:u}{u^Tu}\;\;\\ } \\ }$$ The time-derivative of $\l$ can be expressed in terms of the time-derivative of $H$ $$\large\eqalign{ \dot\l &= \frac{u^T\dot H\:u}{u^Tu} \\ \ \\ }$$