# Derivative of Composite Function/Second Derivative

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## 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 |

$\blacksquare$

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.5$: Permutations and Factorials: Exercise $21$