To make things a little more natural, I will consider the equation $f'(x)=f(x-a)$, $a>0$. As observed in the comments, this is a (linear) delay differential equation. Given a continuous function $\phi\colon[-a,0]\to\mathbb{R}$, there is a unique continuous function $f\colon[-a,\infty)\to\mathbb{R}$, with continuous derivative on $(0,\infty)$, such that $f(x)=\phi(x)$ for $x\in[-a,0]$ and $f'(x)=f(x-a)$ for $x>0$.
The solution $f$ is constructed recursively on the intervals $[na,(n+1)a]$, $n=0,1,2\dots$
If $x\in[0,a]$, then $ f(x)=f(0)+\int_0^xf'(t)dt=f(0)+\int_0^xf(t-a)dt=\phi(0)+\int_{-a}^{x-a}\phi(t)dt. $ If $x\in[a,2a]$, then $ f(x)=f(a)+\int_a^xf'(t)dt=f(a)+\int_a^xf(t-a)dt=f(a)+\int_0^{x-a}f(t)dt, $ which is well defined since in the previous step we calculated $f$ on $[0,a]$. Iterating this procedure, we can find $f$ on any interval $[0,na]$. Only on rare ocasions it is possible to obtain a closed formula for $f$.
Example: $a=1$, $\phi(x)=x^2$.
For $x\in[0,1]$ $ f(x)=\phi(0)+\int_{-1}^{x-1}t^2dt=x-x^2+\frac{x^3}{3}. $ For $x\in[1,2]$ $ f(x)=f(1)+\int_{0}^{x-1}\Bigl(t-t^2+\frac{t^3}{3}. \Bigr)dt=\frac{5}{4} - \frac{7 x}{3} + 2 x^2 - \frac{2 x^3}{3} + \frac{x^4}{12}. $ For $x\in[2,3]$ $ f(x)=f(2)+\int_{1}^{x-1}f(t)dt=\frac{x^5}{60}-\frac{x^4}{4}+\frac{3 x^3}{2}-\frac{13 x^2}{3}+\frac{19 x}{3}-\frac{197}{60}. $ The graph shows the smoothness of $f$.
f (black), $f'$ (red) and $f''$ (blue)]">