$n$-th derivative change of variable I have a change of variables from $\alpha$ to $\beta$ such such that $$\frac{d}{d\beta} = \frac{1}{f(\alpha)}\frac{d}{d\alpha}$$
therfore, the second derivative will read
$$\frac{d^2}{d\beta^2} =\frac{1}{f(\alpha)}\frac{d}{d\alpha}\times \frac{1}{f(\alpha)}\frac{d}{d\alpha}= \frac{1}{f^2(\alpha)}\frac{d^2}{d\alpha^2}-\frac{f'(\alpha)}{f^3(\alpha)}\frac{d}{d\alpha}$$
and the third order reads
$$\frac{d^3}{d\beta^3} =-\frac{f''(\alpha)}{f^4(\alpha)}\frac{d}{d\alpha}+\frac{3 f'^2(\alpha)}{f^5(\alpha)}\frac{d}{d\alpha}-\frac{3 f'(\alpha)}{f^4(\alpha)}\frac{d^2}{d\alpha^2}+\frac{1}{f^3(\alpha)} \frac{d^3}{d\alpha^3}$$
I need to find the $n$-th derivative in $\beta$:
$$\frac{d^n}{d\beta^n} =\, ?$$
is there a general formula for this case?
Yes, known as Leibnitz formula. It gives the $n^{\text{th}} $ derivative of a product
$$(F.G)^{(n)}=\sum_{k=0}^n\binom{n}{k}F^{(k)}G^{(n-k)}$$
Yes, known as Leibnitz formula. It gives the $n^{\text{th}} $ derivative of a product
$$(F.G)^{(n)}=\sum_{k=0}^n\binom{n}{k}F^{(k)}G^{(n-k)}$$