If \( y=\frac{1+x}{1-x} \), find \( \frac{d y}{d x} \)

To find the derivative of \( y = \frac{1+x}{1-x} \), we can apply the quotient rule, which states that if \( y = \frac{u}{v} \), then \[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \] In this case, let \( u = 1 + x \) and \( v = 1 - x \). Now, we calculate the derivatives: - \( u' = \frac{d}{dx}(1+x) = 1 \) - \( v' = \frac{d}{dx}(1-x) = -1 \) Now applying the quotient rule, we get: \[ \frac{dy}{dx} = \frac{(1)(1-x) - (1+x)(-1)}{(1-x)^2} \] Simplifying the numerator: \[ = \frac{1-x + 1 + x}{(1-x)^2} = \frac{2}{(1-x)^2} \] Thus, the derivative is: \[ \frac{dy}{dx} = \frac{2}{(1-x)^2} \]