Function that is the sum of all of its derivatives

$f(x)=\exp(\frac{1}{2}x)$ is such a function, since $f^{(n)}=2^{-n} f(x)$, you have

$$\sum_{n=1}^\infty f^{(n)}(x)=\sum_{n=1}^\infty 2^{-n}f(x)=(2-1)f(x)=f(x)$$

This is the only function (up to a constant prefactor) for which $\sum_{n}f^{(n)}$ and its derivatives converge uniformly (on compacta), as $$f'=\sum_{n=1}^\infty f^{(n+1)}=f-f'$$ follows from this assumption. But this is the same as $f-2 f'=0$, of which the only (real) solutions are $f(x)= C \exp{\frac{x}{2}}$ for some $C \in \mathbb R$.

Differentiate both sides to get $f'(x)=f''(x)+f^{(3)}(x)+...$

So the starting equation becomes $f(x)=f'(x)+f'(x)\Rightarrow f(x)=2f'(x)$

Multiply now both sides by $e^{-\frac{x}{2}}$ and this becomes
$$[e^{-\frac{x}{2}}f(x)]'=0$$
So $f(x)=ce^{\frac{x}{2}}$
Done.

the question means: $$y-y'-y''-y'''-......y^n=0$$ the differential equation is homogeneous and the characteristics equation is $$1-r-r^2-r^3-.......r^n=0$$ $$r(1+r+r^2+r^3+....)=1$$ by using the geometric series (r<1) $$\frac{r}{1-r}=1$$ $$r=1-r$$ $$r=\frac{1}{2}$$ so the function is $$y=Ce^{\frac{x}{2}}$$