## Friday, November 18, 2016

### 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.

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