# Is there a function such that the $n^{th}$ derivative equals the $n^{th}$ power of the first derivative?

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 -

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

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

$$f(x) = -((n-1)!)^{1/(n-1)} \ln(x)$$ is a solution.
• @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. – Brian Tung May 26 at 5:10
• If $y = f'$, the equation says $y^n = y^{(n-1)}$. Constant times a power of $x$ is an obvious thing to try... – Robert Israel May 26 at 14:51