Unizor Derivatives - Cauchy Theorem

Notes to a video lecture on http://www.unizor.com

Derivatives -
Caushy Mean Value Theorem

Cauchy Mean Value Theorem

For two smooth functions f(x) and g(x), defined on segment [a,b] (including endpoints), there exist a point x0∈[a,b] such that the following is true:
f'(x0)/g'(x0) = [f(b)−f(a)]/[g(b)−g(a)]
(with obvious restrictions on denominators not equal to zero).

Proof

Proof of this theorem is based on Rolle's Theorem.
Consider a new function h(x) defined as:
h(x) = f(x) − g(x)·[f(b)−f(a)]/[g(b)−g(a)]

This function satisfies the conditions of Rolle's Theorem:
h(a) = f(a) −g(a)·[f(b)−f(a)]/[g(b)−g(a)] =
= [f(a)·g(b)−f(b)·(g(a)]/[g(b)−g(a)]
h(b) = f(b) − g(b)·[f(b)−f(a)]/[g(b)−g(a)] =
= [f(a)·g(b)−f(b)·(g(a)]/[g(b)−g(a)]

So, h(a) = h(b)

According to Rolle's Theorem, there exists point x0∈[a,b] such that
h'(x0) = 0

Let's find the derivative of function h(x) in terms of derivatives of the original functions f(x) and g(x):
h'(x) = f'(x) − g'(x)·[f(b)−f(a)]/[g(b)−g(a)]
Now the equality to 0 of the derivative of function h(x) at point x0 in terms of original functions f(x) and g(x) looks like this:
0 = f'(x0) − g'(x0)·[f(b)−f(a)]/[g(b)−g(a)]
from which follows
f'(x0)/g'(x0) = [f(b)−f(a)]/[g(b)−g(a)]

End of proof.