Differentiate the following w.r.t. x :
`2l_(n)((x-1)/(x+1))`

Answer

To differentiate the function \( y = 2 \ln\left(\frac{x-1}{x+1}\right) \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the logarithmic function
Using the property of logarithms that states \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \), we can rewrite the function:
\[ y = 2 \left( \ln(x-1) - \ln(x+1) \right) \]

### Step 2: Differentiate using the chain rule
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \left( \frac{d}{dx} \ln(x-1) - \frac{d}{dx} \ln(x+1) \right) \]

Using the derivative of \( \ln(u) \), which is \( \frac{1}{u} \cdot \frac{du}{dx} \), we find:
\[ \frac{d}{dx} \ln(x-1) = \frac{1}{x-1} \cdot \frac{d}{dx}(x-1) = \frac{1}{x-1} \]
\[ \frac{d}{dx} \ln(x+1) = \frac{1}{x+1} \cdot \frac{d}{dx}(x+1) = \frac{1}{x+1} \]

### Step 3: Substitute the derivatives back into the equation
Substituting these derivatives back into the equation gives us:
\[ \frac{dy}{dx} = 2 \left( \frac{1}{x-1} - \frac{1}{x+1} \right) \]

### Step 4: Simplify the expression
Now we need to simplify the expression:
\[ \frac{dy}{dx} = 2 \left( \frac{(x+1) - (x-1)}{(x-1)(x+1)} \right) \]
This simplifies to:
\[ \frac{dy}{dx} = 2 \left( \frac{x + 1 - x + 1}{(x-1)(x+1)} \right) = 2 \left( \frac{2}{(x-1)(x+1)} \right) \]

### Step 5: Final expression
Thus, we have:
\[ \frac{dy}{dx} = \frac{4}{(x-1)(x+1)} = \frac{4}{x^2 - 1} \]

### Final Answer
The derivative of \( y = 2 \ln\left(\frac{x-1}{x+1}\right) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{4}{x^2 - 1} \]