# Faulty proof of chain rule

1. Jan 12, 2010

### jojo12345

I stumbled upon this document that discusses the single variable chain rule:

http://math.rice.edu/~cjd/chainrule.pdf

At the bottom, there is an incorrect proof of the validity of the chain rule, but the author does not cite why the proof is wrong. I'm wondering if the problem is multiplying by g'(x)^-1 without assuming g'(x)~=0 for the relevant x.

2. Jan 12, 2010

### arildno

Your idea is okay, but you are missing the most serious flaws:

a) Consider the second line in the flawed proof.

What interchanges of limiting operation occurs, and why is that interchange illegal?

b) Consider line 3 as well. Can this be regarded as a correct application of the DEFINITION of the derivative?

3. Jan 12, 2010

### jojo12345

I know that if f(x)->a and g(x)->b as x->y, then f(x)g(x)->ab as x->y. So is the problem with the second line that the existence of the limit we are interested in hasn't been established?

4. Jan 12, 2010

### jojo12345

As for the third line, it's certainly not the correct application of the definition of the derivative. However, I don't think it's hard to show that if h(b)->c as b->f(u), then h(f(a))->c as a->u provided f is continuous. So I think that the limit in the third line might still be evaluated correctly.

5. Jan 12, 2010

### elibj123

It's the third line, because it mixes the definition of derivative in respect to the exterior variable (g) and the interior variable (x).

For a function f(g), the derivative f'(g) is

$$lim_{h->0} \frac{f(g+h)-f(g)}{h}$$

Which is generally not identical (even when treating g as a function of x) to:

$$lim_{h->0} \frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}$$

The latter is what written in the third line as the derivative of f wrt g, yet it's not true by definition.

6. Jan 12, 2010

### jojo12345

If
$$f:U\rightarrow\mathbb{R}$$ where $$U$$ is an open neighborhood of $$g(x)$$, then the derivative at $$g(x)$$ can also be defined by $$f^\prime(g(x))=\text{lim}_{b\rightarrow g(x)} \frac{f(b)-f(g(x))}{b-g(x)}$$. Now let $$h(b)=\frac{f(b)-f(g(x))}{b-g(x)}$$. By assumption, $$\text{lim}_{b\rightarrow g(x)} h(b)$$ exists. Call it $$\alpha$$.

Now the third line is saying $$\text{lim}_{u\rightarrow x}h(g(u))=\alpha$$. I claim that this is true because $$g$$, being differentiable at $$x$$, is continuous at $$x$$:

proof: Choose a neighborhood $$U\subseteq\mathbb{R}$$ of $$\alpha$$. There is a neighborhood $$V$$ of $$g(x)$$ such that $$h(V)\subseteq U$$. Now because $$g(x)$$ is continuous at $$x$$ and $$V$$ is an open neighborhood of $$g(x)$$, there is an open neighborhood $$W$$ of $$x$$ such that $$g(W)\subseteq V$$. Thus, $$h(g(W))\subseteq U$$.

7. Jan 12, 2010

### jojo12345

I'm sorry, but the proof I gave in the preceding post is incorrect. Please disregard my previous post.

8. Jan 12, 2010

### jojo12345

The proof would only be correct if $$f(x)$$ mapped into the neighborhood of $$\alpha$$.