Faà di Bruno's Formula/Lemma 2

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:

$\ds \map {D_x} {\prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } } = \prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}$

Proof

\(\ds \)

\(\)

\(\ds \map {D_x} {\prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } }\)

\(\ds \)

\(=\)

\(\ds \prod_{m \mathop = 1}^r \paren {\dfrac 1 {k_m! \paren {m!}^{k_m} } } \map {D_x} {\prod_{m \mathop = 1}^r \paren {\paren {D_x^m u}^{k_m} } }\)

\(\ds \)

\(=\)

\(\ds \prod_{m \mathop = 1}^r \paren {\dfrac 1 {k_m! \paren {m!}^{k_m} } } \sum_{m \mathop = 1}^r \paren {\map {D_x} {\paren {D_x^m u}^{k_m} } \prod_{s \mathop \ne m} \paren {\paren {D_x^s u}^{k_s} } }\)

Product Rule for Derivatives: General Result

\(\ds \)

\(=\)

\(\ds \prod_{m \mathop = 1}^r \paren {\dfrac 1 {k_m! \paren {m!}^{k_m} } } \sum_{m \mathop = 1}^r \paren {\paren {k_m \paren {D_x^m u}^{k_m - 1} D_x^{m + 1} u} \prod_{s \mathop \ne m} \paren {\paren {D_x^s u}^{k_s} } }\)

Faà di Bruno's Formula: Lemma 1

\(\ds \)

\(=\)

\(\ds \prod_{m \mathop = 1}^r \paren {\dfrac 1 {k_m! \paren {m!}^{k_m} } } \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u} \prod_{m \mathop = 1}^r \paren {\paren {D_x^m u}^{k_m} }\)

\(\ds \)

\(=\)

\(\ds \prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}\)

$\blacksquare$