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 $0$ ?

Here is an intuition(!): Let us write $D = \frac{d}{dx}$. If $f$ is analytic,
$$ f(x+h) = \sum_{n=0}^{\infty} \frac{h^{n}}{n!} D^{n} f(x) = e^{hD} f(x) $$
at least in a formal sense (and a genuine sense in view of functional-analysis context). So if you want to solve the equation 
$$ Df(x) = f(x-1) = e^{-D}f(x), $$
you are tempted to solve the equation $D = e^{-D}$, or equivalently, $De^{D} = 1$. Since an eigenfunction of $D$ corresponding to an eigenvalue $\lambda$ is $e^{\lambda x}$, passing to this eigenfunction, it amounts to solve the equation $\lambda e^{\lambda} = 1$.
