How to find the derivative of LNA * LNB * LNC?

How to find the derivative of LNA * LNB * LNC?

How can I remember? It seems to be (1 / a) * LNB * LNC + LNA * (1 / b) * LNC + LNA * LNB * (1 / C)

Can ln (1 + x) be decomposed into LNA + LNB? How to decompose it

Obviously not

How to prove inequality by Lagrange mean value theorem? Could you give me an example

The key is to use "being" ξ ∈ [a, b], so that f '（ ξ)* (B-A) = f (b) - f (a) ", ξ The specific value of F is often unknown, so f '（ ξ) For example, when lsinx sinyl < = LX YL is proved, lsinx sinyl = l (X-Y) cos ξ L < = LX YL. Limited number of words

How to prove the following inequalities with Lagrange mean value theorem 1：larctan a-arctan bl

Consider function   y=arctanx     From Lagrange mean value theorem on [b, a}:
[arctan a-arctan b]/(a-b)=1/(1+ ξ^ 2)≤1
∴|arctan a-arctan b|≤|a-b|

Please prove the inequality with Lagrange mean value theorem  

(1) Let f (x) = e ^ X
Not equal to 0 for any B
According to the mean value theorem, there is u, which satisfies that u is between B and 0, so that (f (b) - f (0)) / (B-0) = f '(U)
Obviously, f '(U) = e ^ u > 1 - > (f (b) - f (0)) / (B-0) > 1 - > F (b) > b + 1 - > e ^ b > b + 1
(2) The second inequality can be obtained from (1), and the first inequality is proved as follows:
Let g (x) = (1 + x) * ln (1 + x)
For any b > 0
According to the mean value theorem, there exists V, which satisfies 01
(g(b)-g(0))/(b-0)=(1+b)*ln(1+b)/b>1 -> ln(1+b)>b/(1+b)

Who knows how Lagrange mean value theorem proves inequalities and identities?

First, to prove the inequality, first set a function related to the problem setting, and then express the formula of Lagrange's mean value theorem. Then, according to the selected value, it must be a limiting condition in the definition domain of the problem setting. Generally, to prove the equality is to set the function in Lagrange's mean value theorem as a function related to the problem setting