Solve $\frac{d}{dx}f(x)=f(x-1)$ I am trying to find a function such that 
$\dfrac{d}{dx}f(x)=f(x-1)$ 
Is there such function other than $0$ ?

If $f : [0;1]$ is any smooth function with $f^{(k+1)}(1) = f^{(k)}(0)$ for all $k$, then we can extend it to a function $f : \Bbb R \to \Bbb R$ by induction with $f(x) = f(1) + \int_0^{x-1} f(t)dt$ if $x \ge 1$ and $f(x) = f'(x+1)$ if $x \le 0$. This new function is a solution to $f'(x) = f(x-1)$.
Conversely, if $f$ is a solution, its restriction to $[0;1]$ is smooth and satifies $f^{(k+1)}(1) = f^{(k)}(0)$ for all $k$, and $f$ is the extension of its restriction. So you actually have a bijection between solutions and smooth functions on $[0;1]$ satisfying $f^{(k+1)}(1) = f^{(k)}(0)$ for all $k$.
