1. **State the problem:** Differentiate the function $$f(x) = \frac{x}{1-x} - \frac{3x^3}{4}$$ with respect to $$x$$. 2. **Differentiate the first term:** Use the quotient rule for $$\frac{x}{1-x}$$. The quotient rule states: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ Here, $$u = x$$ and $$v = 1 - x$$. Calculate derivatives: $$\frac{du}{dx} = 1$$ $$\frac{dv}{dx} = -1$$ Apply quotient rule: $$\frac{d}{dx}\left(\frac{x}{1-x}\right) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1 - x + x}{(1 - x)^2} = \frac{1}{(1 - x)^2}$$ 3. **Differentiate the second term:** $$\frac{d}{dx}\left(\frac{3x^3}{4}\right) = \frac{3}{4} \cdot 3x^2 = \frac{9x^2}{4}$$ 4. **Combine the results:** $$\frac{d}{dx}f(x) = \frac{1}{(1 - x)^2} - \frac{9x^2}{4}$$ **Final answer:** $$\boxed{\frac{d}{dx}\left(\frac{x}{1-x} - \frac{3x^3}{4}\right) = \frac{1}{(1 - x)^2} - \frac{9x^2}{4}}$$