The only differentiable function $f \colon \mathbb R \to \mathbb R$ such that $f^\prime(x)=f(x)$ is $f(x)=ce^{x}$ Prove that the only differentiable function $f \colon \mathbb R \to \mathbb R$ such that $f^\prime(x)=f(x) \mspace{1ex} \forall x\in \mathbb R$ is $f(x)=ce^{x}, \forall x\in \mathbb R$, and for some $c\in \mathbb R$.

Let $g:\mathbb R \to \mathbb R$, $g(x)=f(x)/e^{x}$. Then $$g^\prime(x)=\frac{e^{x}f^\prime(x)-f(x)e^{x}}{e^{2x}}=\frac{f^\prime(x)-f(x)}{e^{x}};$$

by hypothesis $f^\prime(x)=f(x)$ hence $g^\prime(x)=0$, and therefore $g(x)=c$ for some $c\in \mathbb R$.

Then $f(x)=ce^{x}, \forall x\in \mathbb R$, and for some $c\in \mathbb R$.

I would like you to tell me if this proof is correct thank you :)
The proof you have provided is good.
$f(x) = 0$

Is also a function with this property.