If I have two function $f$ and $g$ with $f(x)^{g(x)}$, with $\text{dom}\ f(x)=\{x \in \Bbb R \ \mid f(x)>0\}$ and $g$ defined into your domain, we know that:
$$\psi(x)=f(x)^{g(x)}=e^{g(x)\ln f(x)}\tag 1$$
But developing Taylor's series, it is true that $$e^x=1+x+x^2/2+x^3/6+x^4/24+x^5/120+\dotsb \ \tag 2$$
For the functions of the tipologies $\psi(x)$ I expand using the $(2)$ ,
$$e^{h(x)}=1+h(x)+(h(x))^2/2+\dotsb \tag 3$$
where $h(x)=g(x)\ln f(x) \ $ or are there other ways to expand, easily, $\psi(x)$ from $(1)$?
$$f(x)^{g(x)}=\exp\left(g(x)\ln(f(x))\right)$$ $$h(x):=g(x)\ln(f(x))$$ $$e^{h(x)}=e^{h(0)}\sum_{n=0}^{\infty}B_n(h'(0),h''(0),...,h^{(n-1)}(0))\frac{x^n}{n!}$$
Where $B_n(a_1,...,a_n)$ is the complete Bell polynomial
In your case $$\frac{\mathrm{d}^n}{\mathrm{d}x^n}h(x)=\frac{\mathrm{d}^n}{\mathrm{d}x^n}g(x)\ln(f(x))=\sum_{k=0}^{n}\binom{n}{k}g^{(n-k)}(x)\frac{\mathrm{d}^k}{\mathrm{d}x^k}\ln(f(x))$$
$$\frac{\mathrm{d}^k}{\mathrm{d}x^k}\ln(f(x))=\sum_{j=1}^{k}(-1)^{k-1}\frac{(k-1)!}{f(x)^k}B_{k,j}(f'(x),...,f^{(k-j+1)}(x))$$
Where $B_{k,j}(a_1,...,a_{k-j+1})$ is the incomplete exponential Bell polynomial