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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.

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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. –  Nikita Evseev 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
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$$\frac{\mathrm d^n}{\mathrm dz^n}f(\log\,z)=z^{-n}\sum_{k=1}^n (-1)^{n-k} \left[{n}\atop{k}\right]f^{(k)}(\log\,z)$$ –  J. M. Nov 18 '12 at 13:22

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Set $x=e^t$. Then $t=\log x, f(\log x)=f(t)$ and: $$\frac{d}{dx}=\frac{d}{dt}\cdot\frac{dt}{dx}=\frac{1}{x}\cdot\frac{d}{dt}=e^{-t}\frac{d}{dt},$$ 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 $\{p_n(x)\}_{n\in\mathbb{N}}$ such that: $$ p_1(x)=x,\qquad p_{n+1}(x)= (x-n)\cdot p_n(x),$$ you clearly have: $$ \frac{d^n}{dx^n}f(\log x) = e^{-nt}\cdot \left(p_n\left(\frac{d}{dt}\right)f(t)\right), $$ so: $$ \frac{d^n}{dx^n}f(\log x) = \frac{1}{x^n}\sum_{j=0}^{n-1}(-1)^{j}\cdot e_j(1,\ldots,n-1)\cdot f^{(n-j)}(\log x),$$ where $e_j$ is the $j$-th elementary symmetric polynomial.

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A pretty easy induction shows:

$$\frac{d^k}{dx^k}\{f(\log(x)\}=\left(\left(P_k\left(\frac{d}{dx}\right)f\right)\circ\log\right)(x)\cdot x^{-k} $$

where the polynomial $P_k$ is given by

$$P_k(x):=\prod_{j=0}^{k-1}(x-j) $$

Hopefully this formula is something like what you are after.

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