If $n\in N$ and $f(x)=\ln(1+x^{2n})$, then derivative $f^{(2n)}(-1)=0$.

Asked Apr 06 '16 at 22:05

Active Apr 06 '16 at 22:05

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If $n\in N$ and $f(x)=\ln(1+x^{2n})$, then derivative $f^{(2n)}(-1)=0$.
I try: $e^{f(x)}=1+x^{2n}$ ,$(f'e^f)'=f''e^f+(f')^2e^f$ but I don't know what next.

asked Apr 06 '16 at 22:05

piteer

6,310

Perhaps try induction? The derivatives should have some "regular" form. – Apr 07 '16 at 00:04
That's the only try! – Mhenni Benghorbal Apr 07 '16 at 00:27

If $n\in N$ and $f(x)=\ln(1+x^{2n})$, then derivative $f^{(2n)}(-1)=0$.

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