Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra, and $X,Y\in \mathfrak{g}$ and also let $\mu:M\to \mathfrak{g^*}$ be moment map($M$ smooth manifold) then prove the following equality
$\frac{d}{dt}\mid_{t=0}<\mu(exp (tY))m,X>=i_{Y^\diamond }d<\mu,X>(m)$
Which $Y^\diamond$ is vector field on $M$ generated by $Y$.($m\in M$)
This is just an application of the chain rule which is obscured by the notation. First, note that $\langle \mu(m), X \rangle$ isn't an inner product or anything, it is just the natural pairing between $\mathfrak{g}^\ast$ and $\mathfrak{g}$. In other words, $$\langle \mu(m), X \rangle = \mu(m)(X).$$ The derivation of the identity becomes more transparent when we write the left-hand side in the latter notation: $$\left. \frac{d}{dt}\right|_{t = 0} \langle \mu(\exp(tY)m), X \rangle \quad \leftrightarrow \quad \left. \frac{d}{dt}\right|_{t = 0} \mu(\exp(tY)m)(X).$$ We see that we are taking the derivative of the composite of the following two functions: $$\exp(\cdot Y)m: \Bbb R \longrightarrow M,\, t \mapsto \exp(tY)m, \quad \mu: M \longrightarrow \mathfrak{g}^\ast, \,m\mapsto \mu(m).$$ Hence by the chain rule we have \begin{align*} \left. \frac{d}{dt}\right|_{t = 0} \mu(\exp(tY)m) & = \left. d\mu_m \circ \left(\frac{d}{dt} \exp(tY)m\right) \right|_{t = 0} \\ & = d\mu_m(Y^\diamond(m)). \end{align*} So the identity is $$\left. \frac{d}{dt}\right|_{t = 0} \mu(\exp(tY)m)(X) = (\iota_{Y^\diamond} d\mu)_m(X).$$