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 $g(x)\ne f(x)$?

$$
f = \frac{df}{dx}
$$
$$
dx = \frac{df} f
$$
$$ x + \text{constant} = \log_e |f|
$$
$$
e^x \cdot\text{positive constant} = |f|
$$
$$
e^x \cdot\text{constant} = f.
$$
(One must check separately that $f=0$ is a solution.)
Suppose there were some other solution $g$ (maybe equal to $0$ at some points and not at others?  or whatever $\ldots$).  Then
$$
\underbrace{\frac d {dx}\ \frac{g(x)}{e^x} = \frac{e^x g'(x) - g(x)e^x}{e^{2x}}}_\text{quotient rule}\  \  \underbrace{ = \frac{e^xg(x) - e^x g(x)}{e^{2x}}}_{\text{since }g\,'\,=\,g} = 0,
$$
so $\dfrac{g(x)}{e^x} = \text{constant}$, and we can rule out other solutions.
