# Are there any other functions that behave the same as $ce^x$ with respect to differentiation

$$\frac{d}{dx} ce^x = ce^x$$ Are there any other functions $f$ such that $$\frac{d}{dx} f(x) = f(x)$$ or is $f(x) = ce^x$ the only one?

• Trivial function $f(x)=0$ also works. – John Habert Mar 11 '14 at 3:55
• That corresponds to $c=0$. – vadim123 Mar 11 '14 at 3:55
• The answer depends on your domain... If the domain is disconected, the answer is no ;) – N. S. Mar 11 '14 at 4:15
• @N.S. Can you elaborate on this? – Cameron Martin Mar 11 '14 at 4:17
• If your domain is $(-\infty,0) \cup (0, \infty)$ for example, on each of these intervals you get a function of your type, but the constants can be different. – N. S. Mar 11 '14 at 4:18

No: If $f$ were such a function, consider $g(x) = f(x) e^{-x}$. Then
$$g'(x) = f'(x) e^{-x} + f(x) (-e^{-x}) = f(x) e^{-x} - f(x) e^{-x} = 0$$
As a result, $g$ is constant.
Consider the first order differential equation $$f'(x)=f(x)$$ which is separable. It's integration leads to $$f(x)=c e^x$$ and this is the only possible solution.