# Derivative of Composite Function/Third Derivative

## Theorem

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

Then:

$D_x^3 w = D_u^3 w \paren {D_x^1 u}^3 + 3 D_u^2 w D_x^2 u D_x^1 u + D_u^1 w D_x^3 u$

## 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:

 $\ds D_x^3 w$ $=$ $\ds \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d x^2} }$ $\ds$ $=$ $\ds \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d u^2} \paren {\dfrac {\d u} {\d x} }^2 + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2} }$ Derivative of Composite Function: Second Derivative $\ds$ $=$ $\ds \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d u^2} } \paren {\dfrac {\d u} {\d x} }^2 + \dfrac {\d^2 w} {\d u^2} \map {\dfrac \d {\d x} } {\paren {\dfrac {\d u} {\d x} }^2}$ Product Rule for Derivatives $\ds$  $\, \ds + \,$ $\ds \map {\dfrac \d {\d x} } {\dfrac {\d w} {\d u} } \dfrac {\d^2 u} {\d x^2} + \dfrac {\d w} {\d u} \map {\dfrac \d {\d x} } {\dfrac {\d^2 u} {\d x^2} }$ $\ds$ $=$ $\ds \dfrac {\d^3 w} {\d u^3} \dfrac {\d u} {\d x} \paren {\dfrac {\d u} {\d x} }^2 + \dfrac {\d^2 w} {\d u^2} 2 \dfrac {\d u} {\d x} \dfrac {\d^2 u} {\d x^2}$ Derivative of Composite Function $\ds$  $\, \ds + \,$ $\ds \dfrac {\d^2 w} {\d u^2} \dfrac {\d u} {\d x} \dfrac {\d^2 u} {\d x^2} + \dfrac {\d w} {\d u} \paren {\dfrac {\d^3 u} {\d x^3} }$ $\ds$ $=$ $\ds \dfrac {\d^3 w} {\d u^3} \paren {\dfrac {\d u} {\d x} }^3 + 2 \dfrac {\d^2 w} {\d u^2} \dfrac {\d^2 u} {\d x^2} \dfrac {\d u} {\d x}$ simplifying $\ds$  $\, \ds + \,$ $\ds \dfrac {\d^2 w} {\d u^2} \dfrac {\d^2 u} {\d x^2} \dfrac {\d u} {\d x} + \dfrac {\d w} {\d u} \paren {\dfrac {\d^3 u} {\d x^3} }$ $\ds$ $=$ $\ds \dfrac {\d^3 w} {\d u^3} \paren {\dfrac {\d u} {\d x} }^3 + 3 \dfrac {\d^2 w} {\d u^2} \dfrac {\d^2 u} {\d x^2} \dfrac {\d u} {\d x} + \dfrac {\d w} {\d u} \paren {\dfrac {\d^3 u} {\d x^3} }$ simplifying $\ds$ $=$ $\ds D_u^3 w \paren {D_x^1 u}^3 + 3 D_u^2 w D_x^2 u D_x^1 u + D_u^1 w D_x^3 u$ Leibniz's Notation

$\blacksquare$