If y = log ≤ft( (1 - x^2)/(1 + x^2) ), then dy/dx is equal to

The correct answer is Option (2) → $\frac{-4x}{1 - x^4}$ ##

We have, $y = \log \left( \frac{1 - x^2}{1 + x^2} \right)$

$⇒y = \log(1 - x^2) - \log(1 + x^2) \quad \left[ ∵\log \frac{m}{n} = \log m - \log n \right]$

On differentiating w.r.t. $x$, we get

$\frac{dy}{dx} = \frac{d}{dx} \log(1 - x^2) - \frac{d}{dx} \log(1 + x^2)$

$⇒\frac{dy}{dx} = \frac{1}{(1 - x^2)} \frac{d}{dx}(1 - x^2) - \frac{1}{(1 + x^2)} \frac{d}{dx}(1 + x^2) \quad \left[ ∵\frac{d}{dx} \log x = \frac{1}{x} \right]$

$⇒\frac{dy}{dx} = \frac{-2x}{1 - x^2} - \frac{2x}{1 + x^2}$

$⇒\frac{dy}{dx} = -2x \left[ \frac{1}{1 - x^2} + \frac{1}{1 + x^2} \right] ⇒ \frac{dy}{dx} = -2x \left[ \frac{(1 + x^2) + (1 - x^2)}{1 - x^4} \right]$

$∴\frac{dy}{dx} = \frac{-2x \cdot 2}{1 - x^4} = \frac{-4x}{1 - x^4}$