Derivative of Composite Function/Second Derivative

Theorem

Let $D_x^k u$ denote the $k$th derivative of a function $u$ with respect to $x$.

Then:

$D_x^2 w = D_u^2 w \paren {D_x^1 u}^2 + D_u^1 w D_x^2 u$

Proof

For ease of understanding, let Leibniz's notation be used:

$\dfrac {\d^k u} {\d x^k} := D_x^k u$

Then we have:

 $\ds D_x^2 w$ $=$ $\ds \map {\dfrac \d {\d x} } {\dfrac {\d w} {\d x} }$ $\ds$ $=$ $\ds \map {\dfrac \d {\d x} } {\dfrac {\d w} {\d u} \dfrac {\d u} {\d x} }$ Derivative of Composite Function $\ds$ $=$ $\ds \dfrac {\d u} {\d x} \map {\dfrac {\d} {\d x} } {\dfrac {\d w} {\d u} } + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2}$ Product Rule for Derivatives $\ds$ $=$ $\ds \dfrac {\d u} {\d x} \map {\dfrac {\d} {\d u} } {\dfrac {\d w} {\d u} } \dfrac {\d u} {\d x} + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2}$ Derivative of Composite Function $\ds$ $=$ $\ds \map {\dfrac {\d^2 w} {\d u^2} } {\dfrac {\d u} {\d x} }^2 + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2}$ $\ds$ $=$ $\ds D_u^2 w \paren {D_x^1 u}^2 + D_u^1 w D_x^2 u$ Definition of Leibniz's Notation for Derivatives

$\blacksquare$