Is $f(a+b\epsilon)=f(a)+b\epsilon f'(a)$ true for non-analytic smooth functions of dual argument

Question

Does $f(a+b\epsilon)=f(a)+b\epsilon f'(a)$ remain true for non-analytic smooth functions of dual number argument?
Where $ \epsilon^2=0$, $\epsilon \neq 0$ and $a,b \in \mathbb R$. I found proofs based only on Taylor series. That means it is okay to use the formula for argument values where function is analytic (in real numbers). But what about other cases?

1. Does a proof of $f(a+b\epsilon)=f(a)+b\epsilon f'(a)$ for non-analytic smooth functions exist?

2. What about dual values $b\epsilon g(a)$ that correspond to argument $a$ where function is non-analytic?

Answer 1

For an analytic function of a real input, it's very natural to use the power series to define it on arbitrary dual numbers, and then your result about "automatic differentiation" falls out immediately.

But for a function that's not analytic, it's not generally obvious how to define the function on nonreal inputs, and whether or not you have a similar result depends on how you choose to extend the smooth function to the dual numbers.