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.
How to calculate $\frac{\partial^nf(\ln(x))}{\partial x^n}$
-
1$\frac{\mathrm d^n}{\mathrm dz^n}f(\log\,z)=z^{-n}\sum_{k=1}^n (-1)^{n-k} \left[{n}\atop{k}\right]f^{(k)}(\log\,z)$ – 2012-11-18
2 Answers
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.
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.