用MMA给出复合函数高阶导数公式

青青子衿

复合函数高阶导数公式
Faà di Bruno's formula - Wikipedia
en.wikipedia.org/wiki/Faà_di_Bruno%27s_formula
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1. D[r[t[s]], {s, #}] & /@ Range@4 /. t[s] -> t /.
2.    Derivative[n_][y_][x_] -> Dt[y, {x, n}] //
3.   Column // TraditionalForm

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\begin{align*}
\dfrac{\mathrm{d}r}{\mathrm{d}s}
&=\dfrac{\mathrm{d}r}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}t}{\mathrm{d}s}\\
\dfrac{\mathrm{d}^2r}{\mathrm{d}s^2}
&=\dfrac{\mathrm{d}^2r}{\mathrm{d}t^2}\cdot\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\right)^2+\dfrac{\mathrm{d}r}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}^2t}{\mathrm{d}s^2}\\
\dfrac{\mathrm{d}^3r}{\mathrm{d}s^3}
&=\dfrac{\mathrm{d}^3r}{\mathrm{d}t^3}\cdot\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\right)^3+3\cdot\dfrac{\mathrm{d}^2r}{\mathrm{d}t^2}\cdot\dfrac{\mathrm{d}t}{\mathrm{d}s}\cdot\dfrac{\mathrm{d}^2t}{\mathrm{d}s^2}+\dfrac{\mathrm{d}r}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}^3t}{\mathrm{d}s^3}\\
\dfrac{\mathrm{d}^4r}{\mathrm{d}s^4}
&=\dfrac{\mathrm{d}^4r}{\mathrm{d}t^4}\cdot\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\right)^4+6\cdot\dfrac{\mathrm{d}^3r}{\mathrm{d}t^3}\cdot\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\right)^2\cdot\dfrac{\mathrm{d}^2t}{\mathrm{d}s^2}+\dfrac{\mathrm{d}^2r}{\mathrm{d}t^2}\cdot\left(3\cdot\left(\dfrac{\mathrm{d}^2t}{\mathrm{d}s^2}\right)^2+4\cdot\dfrac{\mathrm{d}t}{\mathrm{d}s}\cdot\dfrac{\mathrm{d}^3t}{\mathrm{d}s^3}\right)+\dfrac{\mathrm{d}r}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}^4t}{\mathrm{d}s^4}\\
\end{align*}
The Faa di Bruno formula revisited
\begin{align*}
\dfrac{\mathrm{d}r}{\mathrm{d}t}
&=\dfrac{\mathrm{d}r}{\mathrm{d}s}\cdot\dfrac{\mathrm{d}s}{\mathrm{d}t}\\
\dfrac{\mathrm{d}^2r}{\mathrm{d}t^2}
&=\dfrac{\mathrm{d}^2r}{\mathrm{d}s^2}\cdot\left(\dfrac{\mathrm{d}s}{\mathrm{d}t}\right)^2+\dfrac{\mathrm{d}r}{\mathrm{d}s}\cdot\dfrac{\mathrm{d}^2s}{\mathrm{d}t^2}\\
\dfrac{\mathrm{d}^3r}{\mathrm{d}t^3}
&=\dfrac{\mathrm{d}^3r}{\mathrm{d}s^3}\cdot\left(\dfrac{\mathrm{d}s}{\mathrm{d}t}\right)^3+3\cdot\dfrac{\mathrm{d}^2r}{\mathrm{d}s^2}\cdot\dfrac{\mathrm{d}s}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}^2s}{\mathrm{d}t^2}+\dfrac{\mathrm{d}r}{\mathrm{d}s}\cdot\dfrac{\mathrm{d}^3s}{\mathrm{d}t^3}\\
\dfrac{\mathrm{d}^4r}{\mathrm{d}t^4}
&=\dfrac{\mathrm{d}^4r}{\mathrm{d}s^4}\cdot\left(\dfrac{\mathrm{d}s}{\mathrm{d}t}\right)^4+6\cdot\dfrac{\mathrm{d}^3r}{\mathrm{d}s^3}\cdot\left(\dfrac{\mathrm{d}s}{\mathrm{d}t}\right)^2\cdot\dfrac{\mathrm{d}^2s}{\mathrm{d}t^2}+\dfrac{\mathrm{d}^2r}{\mathrm{d}s^2}\cdot\left(3\cdot\left(\dfrac{\mathrm{d}^2s}{\mathrm{d}t^2}\right)^2+4\cdot\dfrac{\mathrm{d}s}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}^3s}{\mathrm{d}t^3}\right)+\dfrac{\mathrm{d}r}{\mathrm{d}s}\cdot\dfrac{\mathrm{d}^4s}{\mathrm{d}t^4}\\
\end{align*}

青青子衿

回复 1# 青青子衿
\begin{align*}
\begin{pmatrix}
\dfrac{\mathrm{d}r}{\mathrm{d}s} \\
\dfrac{\mathrm{d}^2r}{\mathrm{d}s^2}  \\
\dfrac{\mathrm{d}^3r}{\mathrm{d}s^3} \\
\dfrac{\mathrm{d}^4r}{\mathrm{d}s^4}
\end{pmatrix}
&=\begin{pmatrix}
\dfrac{\mathrm{d}t}{\mathrm{d}s} \\
\dfrac{\mathrm{d}^2t}{\mathrm{d}s^2}&\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\right)^2  \\
\dfrac{\mathrm{d}^3t}{\mathrm{d}s^3}&3\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\cdot\dfrac{\mathrm{d}^2t}{\mathrm{d}s^2}\right)&\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\right)^3 \\
\dfrac{\mathrm{d}^4t}{\mathrm{d}s^4} &\quad3\left(\dfrac{\mathrm{d}^2t}{\mathrm{d}s^2}\right)^2+4\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\cdot\dfrac{\mathrm{d}^3t}{\mathrm{d}s^3}\right)&6\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\right)^2\cdot\dfrac{\mathrm{d}^2t}{\mathrm{d}s^2}&\quad\left(\dfrac{\mathrm{d}t}{\mathrm{d}s}\right)^4
\end{pmatrix}
\begin{pmatrix}
\dfrac{\mathrm{d}r}{\mathrm{d}t} \\
\dfrac{\mathrm{d}^2r}{\mathrm{d}t^2}  \\
\dfrac{\mathrm{d}^3r}{\mathrm{d}t^3} \\
\dfrac{\mathrm{d}^4r}{\mathrm{d}t^4}
\end{pmatrix}
\end{align*}

青青子衿

根据维基百科，它（复合函数高阶导数的系数）与“部分Bell多项式”有关；
然而，这又与“整数分拆”有关。