Solve for $f(x)=f'(x)$ without previous knowledge. 
Solve for $f(x)=f'(x)$ without previous knowledge.

I know it is obviously $f(x)=e^x$, but could you prove this without knowing $\frac d {dx}e^x=e^x$?
And does there exist a $g(x)=g'(x)$ but $g(x)\ne f(x)$?

Another way do to it, let $(a_k)_{k\in\mathbb{N}}\in\mathbb{R}^\mathbb{N}$ with positive radius of convergence $R$. Let define: $$\forall x\in ]-R,R[,f(x)=\sum_{k=0}^{+\infty}a_kx^k.$$
$f$ satisfies the equation if and only if: $$\forall k\in\mathbb{N}_{\geqslant 1},a_{k-1}=ka_k\Leftrightarrow a_k=\frac{a_{k-1}}{k}.$$
If $a_0=1$, then one has: $$f(x)=\sum_{k=0}^{+\infty}\frac{x^k}{k!}.$$
Using D'Alembert's criterion, one checks that $(a_k)_{k\in\mathbb{N}}$ has infinite radius of convergence.
