I know that $f(x) = e^x$ is its own derivative. It is a monotonically increasing function. It seems intuitively plausible to me that there might be a monotonically decreasing function with the same property. Does one exist?
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I came here to this site to ask this question. When I started filling it in, I came across this question, which let me put together the pieces pretty quickly, so I now know my question is pretty trivial: https://math.stackexchange.com/questions/644879/function-is-equal-to-its-own-derivative Even so, I think this is a fun enough little question that having it spelled out directly might help future readers, so I decided to post my question alongside the answer I discovered anyway. â Kevin Jun 23 '20 at 21:49
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$f(x) = -ce^x, c > 0$
This isn't a particularly exciting answer, but it is the correct one. All functions that are their own derivatives are of the form $f(x) = ce^x, c \in \mathbb{R}$, as explained in this question: Prove that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$
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The requirement yields the differential equation dy - ydx = 0, which has the only solutions y = kexp(x), where k is an arbitrary constant, and f(x) = 0 for all x in the domain. Now f(x) = 0 is decreasing and if k < 0, kexp(x) is a decreasing function. and these are the only collection of functions for which fâ(x) = f(x) with f(x) decreasing.
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