# Faà di Bruno's Formula/Lemma 2

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## Theorem

Let $m \in \Z_{\ge 1}$ be a (strictly) positive integer.

Let $k_m \in \Z_{\ge 1}$ also be a (strictly) positive integer.

Let $u: \R \to \R$ be a function of $x$ which is appropriately differentiable.

Then:

$\displaystyle D_x \left({\prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) }\right) = \prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}$

## Proof

 $\ds$  $\ds D_x \left({\prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) }\right)$ $\ds$ $=$ $\ds \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) D_x \left({\prod_{m \mathop = 1}^r \left({\left({D_x^m u}\right)^{k_m} }\right) }\right)$ $\ds$ $=$ $\ds \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r \left({D_x \left({\left({D_x^m u}\right)^{k_m} }\right) \prod_{s \mathop \ne m} \left({\left({D_x^s u}\right)^{k_s} }\right) }\right)$ Product Rule for Derivatives: General Result $\ds$ $=$ $\ds \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r \left({\left({k_m \left({D_x^m u}\right)^{k_m - 1} D_x^{m + 1} u}\right) \prod_{s \mathop \ne m} \left({\left({D_x^s u}\right)^{k_s} }\right)}\right)$ Faà di Bruno's Formula: Lemma 1 $\ds$ $=$ $\ds \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u} \prod_{m \mathop = 1}^r \left({\left({D_x^m u}\right)^{k_m} }\right)$ $\ds$ $=$ $\ds \prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}$

$\blacksquare$