Derivative of Composite Function/Corollary

Theorem

Let $f, g, h$ be continuous real functions such that:

$y = \map f u, x = \map g u$


Then:

$\dfrac {\d y} {\d x} = \dfrac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x} {\d u} } }$

for $\dfrac {\d x} {\d u} \ne 0$.


Proof

\(\ds \frac {\d y} {\d x} \frac {\d x} {\d u}\) \(=\) \(\ds \frac {\d y} {\d u}\) Derivative of Composite Function
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x}{\d u} } }\) dividing both sides by $\dfrac {\d x} {\d u}$

$\blacksquare$


Sources

  • 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: General Rules of Differentiation: $13.13$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.