Let $G$ be a Lie group, $\mathfrak{g}$ its Lie algebra, $\text{Ad}:G\rightarrow GL(\mathfrak{g})$ the adjoint representation of $G$. Then, for $X,Y\in \mathfrak{g}$,
\begin{align*} \left(\frac{d}{dt}\bigg|_{t=0}\text{Ad}(\exp(tX))\right)Y&=\frac{d}{dt}\bigg|_{t=0}\left(\text{Ad}(\exp(tX))Y\right). \end{align*}
I'm just wondering exactly why this equation is true?
All I'm able to say about the situation is that $\text{Ad}(\exp(tX))$ is a curve in $GL(\mathfrak{g})$, thus \begin{align*} \frac{d}{dt}\bigg|_{t=0}\text{Ad}(\exp(tX)) \end{align*} is a tangent vector in $T_{\text{Id}}GL(\mathfrak{g})$.
Any help here would be much appreciated.
The point is that "evaluating in $Y$" is (the restriction of) a linear map, so its derivative coincides with the map itself. This is not specific to a Lie algebra being involve. If $V$ is a vector space, $v\in V$ an element and $c:I\to GL(V)$ is a smooth curve, then the derivative of the curve $I\to V$ defined by $t\mapsto c(t)(v)$ is just $(c'(t))(v)$.