TITLE: Can exponential functions be thought of as eigenfunctions for the derivative operator?
QUESTION [6 upvotes]: I.e. is the function $y=b^{kx}$ an eigenfunction for the derivative operator $\frac{dy}{dx}$, where k is a constant because the derivative of such a function is ${k\ln(b)}b^{kx}$, which is a constant ($k\ln(b)$) times the original function ($b^{kx}$.) Does it even make sense to say this? If it does, are there any other such eigenfunctions for the derivative operator?

REPLY [4 votes]: To find the eigenvalues and eigenfunctions of the derivative operator we need to solve the differential equation
$$\frac{dy}{dx} = \lambda y$$
which has solutions
$$y = ce^{\lambda x}$$
For any constant $c$, the function above is an eigenfunction with eigenvalue $\lambda$.
Now notice that the function $y = b^{k x}$ is an eigenfunction with eigenvalue $k \ln b$ since
$$y = b^{k x} = (e^{\ln b})^{k x} = e^{k \ln b x}$$