Take the 2-minute tour ×  Is there any sophisticated way to find
n-th derivative of $f(\ln(x))$. Function $f \in
C^{\infty}(\mathbb R)$. I tried 2nd and 3rd but I can not see
any pattern.  share|improve this question You know the Faà di
Bruno formula, I presume? –  Ｊ. Ｍ. Nov 18 '12 at 13:10 @J.M.
Yes, I know but this particular problem for Calculus-I and I
hope one can do it without this formula. –  nikita2 Nov 18 '12
at 13:13 Okay then, would Stirling cycle numbers fall under the
purview of this "Calculus I" that you speak of? –  Ｊ. Ｍ. Nov 18
'12 at 13:19 $$\frac{\mathrm d^n}{\mathrm
dz^n}f(\log\,z)=z^{-n}\sum_{k=1}^n (-1)^{n-k}
\left[{n}\atop{k}\right]f^{(k)}(\log\,z)$$ –  Ｊ. Ｍ. Nov 18 '12
at 13:22  2 Answers 2  up vote 1 down vote accepted  Set
$x=e^t$. Then $t=\log x, f(\log x)=f(t)$ and: $$\frac{d}{dx}=\fr
ac{d}{dt}\cdot\frac{dt}{dx}=\frac{1}{x}\cdot\frac{d}{dt}=e^{-
t}\frac{d}{dt},$$ so: $$\frac{d^n}{dx^n}f(\log
x)=\left(e^{-t}\frac{d}{dt}\right)^n f(t).$$ Now, if you define
a sequence of polynomials $\{p