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We are given the function \( g(x) = \ln \left( \frac{x^2 - 1}{x^2 + 1} \right) \), and we know that its derivative \( g'(x) = \frac{Ax}{x^4 - 1} \). Our goal is to calculate the value of \( A \).

### Step 1: Differentiate \( g(x) \)

To differentiate \( g(x) = \ln \left( \frac{x^2 - 1}{x^2 + 1} \right) \), we use the chain rule and the derivative of the natural logarithm \( \ln(u) \), which gives:

\[
\frac{d}{dx} \left[ \ln(u) \right] = \frac{1}{u} \cdot \frac{du}{dx}
\]

Here, \( u = \frac{x^2 - 1}{x^2 + 1} \), so we will differentiate \( u \) using the quotient rule:

\[
\frac{du}{dx} = \frac{(x^2 + 1)(2x) - (x^2 - 1)(2x)}{(x^2 + 1)^2}
\]

Simplifying the numerator:

\[
= 2x(x^2 + 1 - x^2 + 1) = 4x
\]

Thus, we have:

\[
\frac{du}{dx} = \frac{4x}{(x^2 + 1)^2}
\]

Now, applying the chain rule for the derivative of \( g(x) \):

\[
g'(x) = \frac{1}{\frac{x^2 - 1}{x^2 + 1}} \cdot \frac{4x}{(x^2 + 1)^2}
\]

Simplifying:

\[
g'(x) = \frac{4x(x^2 + 1)}{(x^2 - 1)(x^2 + 1)^2}
\]

### Step 2: Compare with the given derivative

We are told that \( g'(x) = \frac{Ax}{x^4 - 1} \). Notice that:

\[
x^4 - 1 = (x^2 - 1)(x^2 + 1)
\]

So we rewrite the expression for \( g'(x) \) as:

\[
g'(x) = \frac{4x}{(x^2 - 1)(x^2 + 1)}
\]

This matches the form of the given derivative \( \frac{Ax}{x^4 - 1} \). Thus, we can equate:

\[
A = 4
\]

### Final Answer:
The value of \( A \) is 4.

Would you like further details or have any other questions?

### Related Questions:
1. How is the quotient rule applied in differentiation?
2. What is the chain rule in calculus, and how does it apply to logarithmic functions?
3. How can logarithmic differentiation be used in simplifying complex derivatives?
4. What is the importance of simplifying expressions before differentiating?
5. How do you identify patterns in derivatives to match given forms?

### Tip:
Always simplify the inner functions and apply appropriate rules (like the chain rule or quotient rule) to avoid unnecessary complexity when differentiating.

La deriva de: g(x) = ln((x^2 - 1) / (x^2 + 1)) es g'(x) = (Ax) / (x^4 - 1). Calcula 'A'.

In this problem, we calculate the derivative of g(x) = ln((x^2 - 1)/(x^2 + 1)) using the quotient and chain rules. We also solve for A in g'(x) = (Ax)/(x^4 - 1). The problem demonstrates key calculus concepts, including logarithmic differentiation and algebraic simplification. Suitable for university-level calculus students.

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