Talk:Derivative of Composite Function

This theorem can have better prerequisites than that $f,g,h$ be continuous... Compare Continuous Nowhere Differentiable Function

Also, it might be worthwhile to include the multidimensional version. There, total differentiability is required.

Total diffable are continuous functions whose components' partial derivatives all exist. --Lord_Farin 16:12, 22 October 2011 (CDT)


 * Yep, agreed. Bear in mind that the whole mechanism of calculus has (at the moment) been defined solely from the point of view of classical (real) analysis. We haven't even taken complex calculus on board yet, let alone multidimensional. This is purely because of lack of application (I lack patience with the tedious underpinning of analysis and so I haven't had the motivation to buckle down to it).
 * What we really need to do is set up a multi-section page with transclusions (like I can just about set up in my sleep now, I've done so many of them the last few days) so we can define the results for the various different contexts individually and self-containedly, while at the same time allow a common framework to show the parallels of the various approaches. Or something like that. --prime mover 17:22, 22 October 2011 (CDT)

Alternate proof
I don't think it's immediately obvious in this proof why $g^\prime \left({x}\right) = 0 \implies \delta y = 0$ $\forall \delta y$. I think a nicer proof of the chain rule, without all the cases, is:

Define a function $\phi : f\left({D}\right) \to \R$ such that $\phi \left({z}\right) = \begin{cases} \frac {f \left({z}\right) - f \left({g \left({x}\right)}\right)} {z - g \left({x}\right)} & : z \ne g \left({x}\right)  \\ f^\prime \left({z}\right) & : z = g \left({x}\right) \end{cases}$

Note that $\lim_{z \to g \left({x}\right)} \phi \left({z}\right) = f^\prime \left({g \left({x}\right)}\right)$, ie $\phi$ is continuous at $g \left({x}\right)$, meaning $\phi \circ g$ is continuous at $x$ (as g is also continuous at $x$).

Now $\frac {f \left({g \left({y}\right)}\right) - f \left({g \left({x}\right)}\right)} {y - x} = \begin{cases} \frac {f \left({g \left({y}\right)}\right) - f \left({g \left({x}\right)}\right)} {g \left({y}\right) - g \left({x}\right)} \frac {g \left({y}\right) - g \left({x}\right)} {y - x} & : g \left({y}\right) \ne g \left({x}\right) \\ 0 & : g \left({y}\right) = g \left({x}\right) \end{cases}$, $= \phi \left({g \left({y}\right)}\right) \frac {g \left({y}\right) - g \left({x}\right)} {y - x}$, $\forall y \ne x$.

So $\lim_{y \to x} \frac {f \left({g \left({y}\right)}\right) - f \left({g \left({x}\right)}\right)} {y - x} = \lim_{y \to x} \phi \left({g \left({y}\right)}\right) \frac {g \left({y}\right) - g \left({x}\right)} {y - x} = \phi \left({g \left({x}\right)}\right) g^\prime \left({x}\right) = f^\prime \left({g \left({x}\right)}\right) g^\prime \left({x}\right)$, due to the Algebra of Limits and continuity of $\phi \circ g$ at $x$.

It's shorter, and clearer in my opinion--Mathemagician 21:35, 20 May 2012 (EDT)
 * I'm tempted to agree on first gloss. The variation (i.e., $\delta$) approach never won my heart (by sloppy, poorly explained applications in physics lectures, I have actually come to hate it a bit). --Lord_Farin 11:09, 21 May 2012 (EDT)