I have learned about the commutators, and read this:
$$[A, f(B)] = f'[A,B]+\frac{1}{2}f''([A,B]B+B[A,B])+\frac{1}{3!}f'''([A,B]B^2+B[A,B]+B^2[A,B])+...$$
then Simplified to
$$[A, f(B)] = [A,B](f'+f''B+\frac{1}{2}f'''B^2+...)=[A,B]\frac{df}{dB}$$
I do understand the first two equations, only don't understand is why the series $$(f' + f''B + \frac{1}{2}f''' B^2+...)$$ equals to $$\frac{df}{dB}$$