Cauchy mean value theorem in simple language? I am sorry if this is too simple question, but I am having trouble understanding the point and use of "Cauchy mean value theorem". I understood other basic calculus theorems and their proofs. Can you tell me in simple language the explanation and use of it? 
 A: The Cauchy mean theorem states that if $f,g :I \rightarrow \Bbb R $ are differentiable $\mathcal{C^1}$ functions, given $a,b \in \Bbb R$, with $g(a) \neq g(b)$, exists $c \in (a,b)$ such that $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$$
I can edit a quick proof later, if requested, although any decent analysis book should have it.
The mean value theorem is a corollary of this one, taking $g$ as the identity. Note that Cauchy's theorem is much stronger than the mean value theorem. One could naively think that Cauchy's theorem could be proved using the MVT, by the following: $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{\frac{f(b) - f(a)}{b - a}}{\frac{g(b) - g(a)}{b - a}} = \frac{f'(c)}{g'(c)}$$
for some $c$ between $a$ and $b$. However, that  is incorrect, for you do not have the guarantee that the $c$ is the same for both numerator and denominator. The only think you could conclude is that exists $c,d$ between $a$ and $b$ such that the ratio equals $\dfrac{f'(c)}{g'(d)}$.
One nice application of this is proving one of the versions of L'Hospital's rule, together with the squeeze theorem, and the limit of the composition of functions.
Theorem: Let $f,g: I \rightarrow \Bbb R$ be differentiable $\mathcal{C^1}$ functions, such that $f(x_0) = g(x_0) = 0$ and $g'(x_0) \neq 0$. Then $$\lim_{x \to x_0} \frac{f(x)}{g(x)} = \frac{f'(x_0)}{g'(x_0)}$$
Sketch: Notice that: $$\frac{f(x)}{g(x)} = \frac{f(x) - 0}{g(x) - 0} = \frac{f(x) - f(x_0)}{g(x) - g(x_0)}$$
By Cauchy's theorem, exists $c = c(x)$ between $x$ and $x_0$ such that $$\frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}$$
By the squeeze theorem, we have that $c \to x_0$ as $x \to x_0$. Passing the limit, we obtain $$\lim_{x \to x_0} \frac{f(x)}{g(x)} = \lim_{x \to x_0} \frac{f'(c)}{g'(c)} = \frac{f'(x_0)}{g'(x_0)}$$
where the last step is justified by the limit of composition of functions.  $\qquad~\square~ $
A: What we really want to say is as follows: if $f,g$ are defined over $[a,b]$, then there is a $c \in (a,b)$ such that
$$
\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}
$$
that is, there is a point $c$ for which the ratio of the derivatives of $f$ and $g$ is the same as the ratio of the growth of $f$ to the growth of $g$.  However, the fraction makes no sense when $g(b) - g(a) = 0$, so we cross-multiply to get the slightly more general version
$$
f'(c)(g(b) - g(a)) = g'(c)(f(b) - f(a))
$$
Hopefully that clears things up slightly.
