$$ \begin{equation} \begin{aligned} \exp(\hat{A})=\sum_{m=0}^{\infty}\frac{1}{m!}\hat{A}^{m}. \end{aligned} \end{equation}
$$
$$ \begin{equation} \begin{aligned} \left(\frac{\mathrm{d}}{\mathrm{d}t}\hat{A}^n\right) & = \sum_{k = 0}^{n-1}\hat{A}^k\left(\frac{\mathrm{d}}{\mathrm{d}t}\hat{A}\right)\hat{A}^{n-k-1}. \end{aligned} \end{equation}
$$
Mathematical Methods of Quantum Optics
Sneddon’s formula: [p37]
$$ \begin{equation}\begin{aligned}\frac{\mathrm{d}}{\mathrm{d}t}\exp[\hat{A}(t)]=\int_0^1\exp[u\hat{A}]\frac{\mathrm{d}\hat{A}}{\mathrm{d}t}\exp[(1-u)\hat{A}]\mathrm{d}u.\end{aligned}\end{equation} $$
If $\hat A(t)=\hat A\ t$, as an example, then this yields the familiar c-number result
$$ \frac{\mathrm{d}}{\mathrm{d}t}\exp(\hat{A}t)=\hat{A}\exp(\hat{A}t)=\exp(\hat{A}t)\hat{A}. $$
Consider a spin -1/2, any of its component $\hat S_a$ along a direction $\boldsymbol a$ is a two-dimensional operator which can assume the values $\pm 1/2:\hat S_a^2=\frac 1 4+0\cdot \hat S_a.$
$$ \prod_{i=1}^n\left(\hat{X}-\lambda_i\right)=0\to \hat{X}^N=\sum_{m=0}^{N-1}\alpha_m\hat{X}^m. $$
Note that$C_k(0)=\delta {k0}$, make $C{-1}(\theta)=0$
$$ \sum_{m=1}^NC_{m-1}(\theta)\hat{X}^m=\sum_{m=0}^{N-1}\dot{C}_m(\theta)\hat{X}^m. $$
$$ \dot{C}k(\theta)=\alpha_kC{N-1}(\theta)+C_{k-1}(\theta),\quad C_{-1}(\theta)=0, $$
The solution of spin is
$$ \exp(\theta\hat{S}_a)=\cosh\left(\frac{\theta}{2}\right)+2\sinh\left(\frac{\theta}{2}\right)\hat{S}_a. $$
spin-1:
$$ \exp(\theta\hat{J}_a)=1+\sinh(\theta)\hat{J}_a+(\cosh(\theta)-1)\hat{J}_a^2. $$
Let $\hat S$ be a non-singular operator. The transformation defined by
$$ \hat S^{-1}\hat A\hat S=\hat B $$