Question

Differentiate In(1-x^2/1+x^2)

Answer 1

To differentiate the function \( f(x) = \ln\left(\frac{1 - x^2}{1 + x^2}\right) \), we will use the chain rule and the quotient rule.

1. **Apply the properties of logarithms**:
\[
f(x) = \ln(1 - x^2) - \ln(1 + x^2)
\]

2. **Differentiate each term**:
Using the derivative of \(\ln(u)\), which is \(\frac{1}{u} \cdot u'\):
\[
f'(x) = \frac{d}{dx} \left( \ln(1 - x^2) \right) - \frac{d}{dx} \left( \ln(1 + x^2) \right)
\]

- For \(\ln(1 - x^2)\):
\[
\frac{d}{dx} \ln(1 - x^2) = \frac{1}{1 - x^2} \cdot (-2x) = \frac{-2x}{1 - x^2}
\]

- For \(\ln(1 + x^2)\):
\[
\frac{d}{dx} \ln(1 + x^2) = \frac{1}{1 + x^2} \cdot (2x) = \frac{2x}{1 + x^2}
\]

3. **Combine the results**:
\[
f'(x) = \frac{-2x}{1 - x^2} - \frac{2x}{1 + x^2}
\]

4. **Find a common denominator**:
The common denominator for \(1 - x^2\) and \(1 + x^2\) is \((1 - x^2)(1 + x^2)\):
\[
f'(x) = \frac{-2x(1 + x^2) - 2x(1 - x^2)}{(1 - x^2)(1 + x^2)}
\]

Simplifying the numerator:
\[
-2x(1 + x^2) - 2x(1 - x^2) = -2x - 2x^3 - 2x + 2x^3 = -4x
\]

5. **Final result**:
Therefore, the derivative is:
\[
f'(x) = \frac{-4x}{(1 - x^2)(1 + x^2)} = \frac{-4x}{1 - x^4}
\]

Thus, the derivative of \( f(x) = \ln\left(\frac{1 - x^2}{1 + x^2}\right) \) is:
\[
f'(x) = \frac{-4x}{1 - x^4}
\]