Derivative of function in Ad

Question

If $G$ is a Lie group and $Ad: G \rightarrow End(\mathfrak{g})$ is the adjoint representation, what is the derivative of $Ad(\exp ty)(tH)$ in $t=0$ where $y, H\in\mathfrak{g}$? I know that $ad$ is the derivative of $Ad$ in $t=0$. Thanks in advance.

Answer 1

I'm not sure I appreciate your notation, but since $\operatorname{Ad}_{\exp ty} = e^{t \operatorname {ad}_y }$, you have $$ \operatorname{Ad}_{\exp ty} ~ tH= e^{t \operatorname {ad}_y } tH , $$ so the derivative of this w.r.t. t is $$ e^{t \operatorname {ad}_y } ( t \operatorname{ad}_y ~H+ H), $$ which reduces to H at t=0. Is this what you are unsure about?

If, instead, the original argument were H instead of tH, you'd get ad_y H .