Step 1 : Understanding the Question:
The topic for this problem is Advanced Differentiation, specifically Logarithmic Differentiation. This technique is mandatory when dealing with functions where the variable $x$ appears in both the base and the exponent (i.e., functions of the form $u(x)^{v(x)}$). Direct application of the power rule or the exponential rule is not possible here. By taking the natural logarithm of both sides, we convert the exponent into a multiplier, allowing us to use the product rule and chain rule more easily.
Step 2 : Key Formulas and approach:
1. Logarithmic Differentiation Property: If $y = u^v$, then $\ln y = v \ln u$.
2. Product Rule: $\frac{d}{dx}(u \cdot v) = u v' + v u'$.
3. Chain Rule for Logarithms: $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$.
4. Approach: Take the natural log of both sides, use the property $\ln(a/b) = \ln a - \ln b$ to simplify the base, differentiate both sides with respect to $x$, and solve for $dy/dx$.
Step 3 : Detailed Explanation:
Given the function $y = \left(\frac{x+1}{x-1}\right)^x$, we take the natural logarithm on both sides: $\ln y = \ln \left[\left(\frac{x+1}{x-1}\right)^x\right]$.
Using the log power property, this becomes: $\ln y = x \ln \left(\frac{x+1}{x-1}\right)$.
Further simplifying the log using division properties: $\ln y = x [\ln(x+1) - \ln(x-1)]$.
Now, we differentiate both sides with respect to $x$. On the left side, $\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}$.
On the right side, we apply the product rule to $x$ and $[\ln(x+1) - \ln(x-1)]$.
The derivative is: $\frac{1}{y} \frac{dy}{dx} = (x) \cdot \frac{d}{dx}[\ln(x+1) - \ln(x-1)] + [\ln(x+1) - \ln(x-1)] \cdot \frac{d}{dx}(x)$.
Calculating the derivatives: $\frac{d}{dx}[\ln(x+1) - \ln(x-1)] = \frac{1}{x+1} - \frac{1}{x-1}$.
Combining terms with a common denominator: $\frac{1}{x+1} - \frac{1}{x-1} = \frac{(x-1) - (x+1)}{(x+1)(x-1)} = \frac{-2}{x^2-1}$.
Substituting back into our product rule expression: $\frac{1}{y} \frac{dy}{dx} = x \left(\frac{-2}{x^2-1}\right) + \ln \left(\frac{x+1}{x-1}\right)$.
Rearranging the terms: $\frac{1}{y} \frac{dy}{dx} = \ln \left(\frac{x+1}{x-1}\right) - \frac{2x}{x^2-1}$.
Finally, multiply by $y$ to solve for $dy/dx$: $\frac{dy}{dx} = y \left[\ln\left(\dfrac{x+1}{x-1}\right)-\dfrac{2x}{x^2-1}\right]$.
Step 4 : Final Answer:
Using logarithmic differentiation, the derivative is found to be $y[\ln(\frac{x+1}{x-1})-\frac{2x}{x^2-1}]$, which is Option (A).