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Consider a function , i.e., $f(x)=\frac{a^x}{x}$, where $x>0$, $0<a<1$. Now I want to determine the sign of it's $n$-th derivative, i..e, $f^{(n)}(x)$.

My conclusion is that the sign of $f^{(n)}(x)$ is $(-1)^n$ which holds at least when $n=0,1$.

Can anyone give me some hints to prove that the above conclusion still holds for any $n=0,1,2,\cdots$?

Any comment is appreciated.

Answer

Motivated by @pushpen.paul, I have known that the sign of $f^{(n)}(x)$ is $(-1)^n$. Here is the proof.

Since $f(x)=\frac{a^x}{x}=\int_{0}^a t^{x-1}dt$, we have $f^{(n)}(x)=\int_{0}^a t^{x-1}(\log(t))^ndt$. Due to $0<t<a<1$, the sign of $(\log(t))^n$ is $(-1)^n$, which completes the proof.

Dave
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    You can use the fact that $f(x)=\frac{a^x}{x}=\int_0^a t^{x-1}dt$, hence $f^{(n)}(x)=\int_0^a t^{x-1}(\log t)^ndt$. – Kelenner Nov 15 '17 at 12:44
  • Great! A nice proof! Thank you. – Dave Nov 15 '17 at 12:52
  • @Dave Since you understood the solution, you may consider the solution in your question/post an answer to your question. Just a personal thought :-) – hola Nov 15 '17 at 12:54

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