1. **State the problem:** Differentiate the function $$y = \frac{1+x}{1-x}$$ with respect to $$x$$ and find $$\frac{dy}{dx}$$. 2. **Apply the quotient rule:** For $$y = \frac{u}{v}$$, $$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$. 3. Identify $$u = 1+x$$ and $$v = 1-x$$. 4. Compute derivatives: $$\frac{du}{dx} = 1$$ and $$\frac{dv}{dx} = -1$$. 5. Substitute into quotient rule formula: $$\frac{dy}{dx} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2}$$ 6. Simplify numerator: $$(1-x) + (1+x) = 1-x + 1 + x = 2$$ 7. So derivative is: $$\frac{dy}{dx} = \frac{2}{(1-x)^2}$$ **Answer:** Option c. $$2(1-x)^2$$