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 $$\exists x \in \left({a \, . \, . \, b}\right): g^{\prime} \left({x}\right) \ne 0$$.

Then $$\exists \xi \in \left({a \, . \, . \, b}\right): \frac {f^{\prime} \left({\xi}\right)} {g^{\prime} \left({\xi}\right)} = \frac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$$.

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.

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

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

That is, $$h = \frac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$$.

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

That is, $$h = - \frac {f^{\prime} \left({\xi}\right)} {g^{\prime} \left({\xi}\right)}$$.

Hence, from the definition of the derivative, the result.