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Letting

$$\begin{align*}x&=f(t)\\y&=g(t)\end{align*}$$

the following expressions are well known:

$$\begin{align*}\frac{\mathrm dy}{\mathrm dx}&=\frac{g^\prime (t)}{f^\prime (t)}\\\frac{\mathrm d^2 y}{\mathrm dx^2}&=\frac{f^\prime (t)g^{\prime\prime} (t)-g^\prime (t)f^{\prime\prime} (t)}{f^\prime (t)^3}\end{align*}$$

With some effort, we can derive the expression for the third derivative:

$$\frac{\mathrm d^3 y}{\mathrm dx^3}=\frac{f'(t) \left(g^{(3)}(t) f'(t)-3 f''(t) g''(t)\right)+g'(t) \left(3 f''(t)^2-f^{(3)}(t)f'(t)\right)}{f'(t)^5}$$

After deriving expressions for the next higher derivatives, I am unable to detect any particular pattern in the expressions, save for the denominator $f'(t)^{2n-1}$ of the $n$-th derivative. I've also tried to search around for information on the derivatives of parametrically-defined functions, but no dice.

Here then is my question: is there a general formula for $\dfrac{\mathrm d^n y}{\mathrm dx^n}$ in terms of $f(t),g(t)$ and their derivatives?

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    If $\;\mathrm{d}^ny/\mathrm{d}x^n=h_n(t)f\,'(t)^{1-2n}$ then $h_1=g'$ and $h_{n+1}=h'_nf\,'-(2n-1)h_nf\;''$. I got that much.2011-10-01
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    My instinct says that an elementary formula is unlikely since the $n$-th derivative will be a linear combination of a product of $(n-1)$ derivatives of $f$ (of various degrees) and a derivative of $g$ (of some degree). This means we have a $n$-dimensional array of coefficients for $\dfrac{\mathrm{d}^n y}{\mathrm{d} x^n}$...2011-10-01
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    I guess you could 'factorize' it as $$\frac{d^ny}{dx^n}=\frac{1}{(f')^{2n+1}}\left[\prod_{k=1}^n\left(f'\frac{d}{dt}-(2k-1)f''\right)\right]g'.$$ Edit: It seems I can't make the $d$'s above into \mathrm style in comments, but I can in the answer preview.2011-10-01
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    @Zhen: I know (and I did have the foresight not to say the word "elementary" in the question body); I'm down with an expression involving Bell polynomials or somesuch. ;)2011-10-01
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    Well, actually, perhaps something might be possible if we use the Faà di Bruno formula. After all, in order to divide by $f'(t)$, $f'(t) \ne 0$, and so $f$ is locally invertible by the inverse function theorem. Then we can express $y$ as a function of $x$ (locally) by a composite of $g$ and $f^{-1}$...2011-10-01
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    @anon: your recurrence and "operator product" formulae look good; I wonder if there's a more explicit version of those. The reason why I'm trying to look for an explicit expression is that I have values of $f^{(n)}(t)$ and $g^{(n)}(t)$ for a given value of $t$ available, so I'm hoping for a formula I can plug these values into...2011-10-01
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    @anon it seems to me that you meant $\frac{d^{n+1}y}{dx^{n+1}}=\frac{1}{(f')^{2n+1}}\left[\prod_{k=1}^n\left(f' \frac{d}{dt}-(2k-1)f''\right)\right]g'.$2011-10-01
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    @Emanuele: Yes, that's correct. Now if we take the concept Zhen mentioned and split $$h_n=\sum_{m=1}F_{m,n} g^{(m)},$$ where $F_{0,m}$ and $F_{k,m}$ for $k>m$ are identically $0$, we obtain $$F_{m,n+1}=f'(F_{m-1,n}+F'_{m,n})-(2n-1)f''F_{m,n}$$ Eck.2011-10-01
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    @J.M.: I added some results to my original answer that may interest you. I'm not sure whether you'll be notified when I edit an old answer, so I'm pinging you as well.2012-12-11

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