Problem

Calculate the derivative of the function g(x): $g(x)=\ln \left(\frac{x+1}{x-1}\right)$

Solution

Step 1 :Identify the outer function as \(\ln(u)\) and the inner function as \(u = \frac{x+1}{x-1}\).

Step 2 :Find the derivative of the outer function, which is \(\frac{1}{u}\).

Step 3 :Find the derivative of the inner function using the quotient rule. The derivative of \(\frac{x+1}{x-1}\) is \(\frac{(x-1)*1 - (x+1)*1}{(x-1)^2} = \frac{-2}{(x-1)^2}\).

Step 4 :Apply the chain rule to find the derivative of the function. The derivative of \(g(x)\) is the derivative of the outer function times the derivative of the inner function, which is \(\frac{1}{u} * \frac{-2}{(x-1)^2} = \frac{-2}{(x+1)(x-1)}\).

Step 5 :\(\boxed{g'(x) = \frac{-2}{(x+1)(x-1)}}\) is the derivative of the function \(g(x) = \ln\left(\frac{x+1}{x-1}\right)\).

From Solvely APP
Source: https://solvelyapp.com/problems/NSJEC2JlTj/

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