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Let $y=f^{-1}(x)$. As we know:
\begin{align} \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{{f}'(y)} \end{align} Thereof we have:
\begin{align} \frac{\mathrm{d^2} y}{\mathrm{d} x^2}=\frac{-{f}''(y)}{({f}'(y))^3} \end{align}

\begin{align} \frac{\mathrm{d^3} y}{\mathrm{d} x^3}=\frac{3({f}''(y))^2-{f}'(y){f}'''(y)}{({f}'(y))^5} \end{align} Is there a general rule for
\begin{align} \frac{\mathrm{d^n} y}{\mathrm{d} x^n}=?\end{align}

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  • $\begingroup$ nice question$~~\color{blue}{+1}~~$ $\endgroup$ – Software Aug 27 '13 at 1:28
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One expression is $$y^{(n)}(x_0)=\bigl((y-y_0)(\frac{y-y_0}{f(y)-x_0})^{n+1} f'(y)\bigr)^{(n)}|_{y=y_0}$$ where $y_0=f^{-1}(x_0)$ (it is a constant). Supposing $f$ is holomorphic, it can be derived from the residue theorem: $y^{(n)}(x_0)=n!/2\pi i\,\oint (y-y_0)/(x-x_0)^{n+1} dx =n!/2\pi i\,\oint ((y-y_0)/(x-x_0))^{n+1}/(y-y_0)^{n}f'\,dy$

(the residue theorem is taught as a way of computing integrals in terms of derivatives, but in some cases it is a means of computing derivatives in terms of integrals)

BTW, it becomes more entertaining when $h$ is another function and you want to find the n-th derivative of $h(y(x))$. Supposing $f(0)=0$, and that we compute at $0$ (i.e. $x_0=y_0=0$, we get $$\bigl(h(y)(\frac{y}{f(y)})^{n+1} f'(y)\bigr)^{(n)}|_{y=0}$$ As an example, if $h(y)=1/(1-y)$, $x=ye^{-y}$, we get the Taylor series $$h(y)=\sum_n\frac{n^n}{n!}x^n.$$

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  • $\begingroup$ It seems a recursive solution. By the way, thanks for your answer. Would you please find the 2nd derivative I mentioned in the question with your first formula? I do not know why I can not obtain that? (useful links: en.wikipedia.org/wiki/… and scribd.com/doc/13699758/…). $\endgroup$ – Amir Kazemi Aug 27 '13 at 18:54

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