Chain Rule for Partial Derivatives

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Theorem

Let $F: \R^2 \to \R$ be a real-valued function of $2$ variables.

Let $X: \R^2 \to \R$ and $Y: \R^2 \to \R$ also be real-valued functions of $2$ variables.


Let $F = \map f {x, y}$ be such that:

\(\ds x\) \(=\) \(\ds \map X {u, v}\)
\(\ds y\) \(=\) \(\ds \map Y {u, v}\)


Then:

$F = \map F {u, v}$

and:

\(\ds \dfrac {\partial F} {\partial u}\) \(=\) \(\ds \dfrac {\partial f} {\partial x} \dfrac {\partial X} {\partial u} + \dfrac {\partial f} {\partial y} \dfrac {\partial Y} {\partial u}\)
\(\ds \dfrac {\partial F} {\partial v}\) \(=\) \(\ds \dfrac {\partial f} {\partial x} \dfrac {\partial X} {\partial v} + \dfrac {\partial f} {\partial y} \dfrac {\partial Y} {\partial v}\)



This article is complete as far as it goes, but it could do with expansion.
In particular: Conditions on continuity and/or differentiability need to be incorporated here
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Corollary 1

Let $F = \map f {x, y}$ be a real-valued function from $\R^2$ to $\R$.

Let $x = \map X t$ and $y = \map Y t$ be real functions.


Then:

$F = \map F t$

and:

$\dfrac {\d F} {\d t} = \dfrac {\partial F} {\partial x} \dfrac {\d x} {\d t} + \dfrac {\partial F} {\partial y} \dfrac {\d Y} {\d t}$


Corollary 2

Let $F = \map f {x, y}$ be a real-valued function from $\R^2$ to $\R$.

Let $y = \map Y x$ be a real function.


Then:

$\dfrac {\d F} {\d x} = \dfrac {\partial F} {\partial x} + \dfrac {\partial F} {\partial y} \dfrac {\d Y} {\d x}$


Proof


This theorem requires a proof.
In particular: Discussed here: https://www.ma.imperial.ac.uk/~jdg/AECHAIN.PDF which is appropriate considering this result I'm reading from a reference book written by staff at Imperial.
Use Cauchy-Riemann Equations (a special case) as a basis for this, and make that page dependent upon this.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.
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Sources

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