Cauchy Mean Value Theorem

Theorem
Let $f$ and $g$ be a real functions which are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

Suppose that:
 * $\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$

Then:
 * $\exists \xi \in \left({a \,.\,.\, b}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$

Proof
Let $F$ be the real function defined on $\left[{a \,.\,.\, b}\right]$ by $F \left({x}\right) = f \left({x}\right) + h g \left({x}\right)$, where $h \in \R$ is a constant.

The plan is to choose the constant $h$ such that $F \left({a}\right) = F \left({b}\right)$ and so apply Rolle's Theorem.

We need to make:
 * $f \left({a}\right) + h g \left({a}\right) = f \left({b}\right) + h g \left({b}\right)$

We have that:
 * $\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$

So, by Rolle's Theorem:
 * $g \left({a}\right) \ne g \left({b}\right)$

Thus we can let:
 * $h = - \dfrac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$

So, by Rolle's Theorem:
 * $\exists \xi \in \left({a \,.\,.\, b}\right): 0 = F' \left({\xi}\right) = f' \left({\xi}\right) + h g' \left({\xi}\right)$

That is:
 * $h = - \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)}$

Hence the result, from the definition of the derivative.