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I have seen functions whose second derivative is equal to the square of the first derivative. So is there a function whose $n^{th}$ derivative is equal to the $n^{th}$ power of the first derivative. Or, is there a function such that -

$$\left(\frac{df}{dx}\right)^n=\frac{d^{n}f}{dx^{n}}$$

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    $\begingroup$ this is just an ode.. $y^{(n)} - (y')^n = 0$ $\endgroup$ – K. Y May 26 at 3:24
  • $\begingroup$ Note that any constant function works. $\endgroup$ – K. Y May 26 at 3:27
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    $\begingroup$ $f(x)=0$ The closest other functions I can think of are exponential functions. They don't satisfy the equation, but they're pretty close to. $\endgroup$ – Badr B May 26 at 3:35
  • $\begingroup$ WA can't solve it even for $n=3$... $\endgroup$ – Micah May 26 at 3:52
  • $\begingroup$ Did you want this to be true for all $n$, or just a given $n$? $\endgroup$ – Brian Tung May 26 at 5:00
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$f(x) = -((n-1)!)^{1/(n-1)} \ln(x)$ is a solution.

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  • $\begingroup$ How did you find that? That's quite amazing. $\endgroup$ – Badr B May 26 at 4:28
  • $\begingroup$ Like @BadrB, I too would like at least a general story for how you found this. $\endgroup$ – Brian Tung May 26 at 4:59
  • $\begingroup$ @Somos: I think that is an error in the edit by Parcly Taxel. That previously read $-[(n-1)!]^\frac{1}{n-1}\ln x$, and the $(n-1)!$ was replaced by $\Gamma(x)$, rather than $\Gamma(n)$. (Note that $n$ is presumed to be an integer, so the expression did not need to be made continuous.) I'm going to revert the edit. $\endgroup$ – Brian Tung May 26 at 5:10
  • $\begingroup$ @obertIsrael can you please tell us how you arrived at that solution or did you know this before? $\endgroup$ – NoLand'sMan May 26 at 6:06
  • $\begingroup$ If $y = f'$, the equation says $y^n = y^{(n-1)}$. Constant times a power of $x$ is an obvious thing to try... $\endgroup$ – Robert Israel May 26 at 14:51

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