If y = log(1−x²/1+x²), then dy/dx is equal to

The correct option is −4x/1−x⁴.

To find the derivative of the given function, we first write down the function:

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

We can simplify the expression by using the logarithmic property $\log \left(\frac{a}{b}\right)=\log \left(a\right)-\log \left(b\right)$:

$y=\log (1-x^2)-\log (1+x^2)$

Now, we differentiate both sides with respect to x using the chain rule:

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

Since the derivative of $1-x^2$ is $-2x$ and the derivative of $1+x^2$ is $2x$, we get:

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

Now, we can factor out $-2x$ and find a common denominator:

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

Combine the fractions inside the bracket:

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

Simplify the numerator and denominator:

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