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

$\blacksquare$


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