Differentiate the following w.r.t. x :

• log(1 + x) / (1 - x)

To differentiate the function \( f(x) = \frac{\log(1 + x)}{1 - x} \) with respect to \( x \), we will use the quotient rule and the chain rule. The quotient rule states that if you have a function in the form of \( \frac{u}{v} \), its derivative is given by:

Quotient Rule Formula

If \( u = \log(1 + x) \) and \( v = 1 - x \), then:

\( f'(x) = \frac{u'v - uv'}{v^2} \)

Step 1: Find \( u' \) and \( v' \)

• For \( u = \log(1 + x) \):

  Using the chain rule, \( u' = \frac{1}{1 + x} \cdot (1) = \frac{1}{1 + x} \).

• For \( v = 1 - x \):

  The derivative is \( v' = -1 \).

Step 2: Apply the Quotient Rule

Now substituting \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula:

\( f'(x) = \frac{\left(\frac{1}{1 + x}\right)(1 - x) - (\log(1 + x))(-1)}{(1 - x)^2} \)

Step 3: Simplify the Expression

Now, simplifying the numerator:

• First term: \( \frac{1 - x}{1 + x} \)
• Second term: \( \log(1 + x) \)

Thus, the derivative becomes:

\( f'(x) = \frac{\frac{1 - x}{1 + x} + \log(1 + x)}{(1 - x)^2} \)

Final Result

The derivative of \( f(x) = \frac{\log(1 + x)}{1 - x} \) with respect to \( x \) is:

\( f'(x) = \frac{(1 - x) + (1 + x)\log(1 + x)}{(1 + x)(1 - x)^2} \)