Derivative of Composite Function/Corollary

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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

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

$\blacksquare$


Sources