Is it true that $f^k=(k-1)!/log^k(t)$ ? $k$ refers to the order of derivatives and $f(t)=logt$.If yes, then how can i show this? I've been doing different ways just to solve it but i cant get the right one. Can someone help me?
2026-03-27 10:43:04.1774608184
Is the equality correct?
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No, this is not the case. Notice that
$f'(x)=\frac1x$
$f''(x)=-\frac1{x^2}$
$\vdots$
$f^{(k)}(x)=(-1)^{n+1}\frac{(n-1)!}{x^k}$
So you were close, but not quite there.