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February 16th, 2017, 01:47 AM   #1
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Multivariable Chain rule proof : is it correct?

Hello,

I've had a go at proving the chain rule for a composition that maps from R^2 to R to R. There are two images attached. I've used the "arithmetic of limits" rules throughout, that if two limits exist, their product limit and sum limit exist etc. Any thoughts / scrutiny would be appreciated, since I can't find a proof elsewhere.

James
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File Type: jpg image1.jpg (88.4 KB, 4 views)
File Type: jpg image2.jpg (89.9 KB, 2 views)
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February 16th, 2017, 03:49 AM   #2
SDK
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2 problems with your proof:

1. Look over your computations on page 1 (about halfway down). Consider if $f$ is constant and notice that none of these statements you have made are true. Namely, the RHS of (*) is not true when $f$ is constant. Unfortunately, this computation is the crux of your proof and the rest of the computation falls through. The difficult part of proving the chain theorem occurs precisely when $f$ has zero derivative on some sequence converging to $(a,b)$.

2. On page 2, you state from one line to the next that
$$ \lim_{h \rightarrow 0 } \frac{g(f(a+h,b))-g(f(a,b))}{f(a+h,b) - f(a,b)} = g'(f(a,b))$$

but this is precisely the statement that requires proof. Implicitly, in the line above you have interchanged the limit for the variable you call $t$ with the variable $h$. In this case that can be justified but this isn't always the case and any proof should argue why this is allowed here.

As a hint toward a proof I suggest you consider the following expression
$$
F(h) = \frac{g(f(a+h,b)) - g(f(a,b))}{h} - g'(f(a,b)) \qquad h \neq 0 \\
F(0) = 0.
$$
Now try to prove that $F$ is continuous at $h= 0$. Hope this helps.
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February 16th, 2017, 10:16 AM   #3
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SDK ,

Thanks for the reply, I've attached an image to (try) to build on what you said about the constant f function. As for the rest, I'm going to assume the function is non constant and proceed with your idea.

James
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February 16th, 2017, 10:28 AM   #4
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Also, am I right in saying that to prove the continuity, one would have to use the fact that g is differentiable f(a,b)? Then manipulate the limit?
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