1. **Problem (i):** Given $y = \frac{x^2 - x}{e^x}$, find $\frac{dy}{dx}$ and simplify. 2. **Formula:** Use the quotient rule for derivatives: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$ where $u = x^2 - x$ and $v = e^x$. 3. **Calculate derivatives:** - $\frac{du}{dx} = 2x - 1$ - $\frac{dv}{dx} = e^x$ 4. **Apply quotient rule:** $$\frac{dy}{dx} = \frac{e^x(2x - 1) - (x^2 - x)e^x}{(e^x)^2}$$ 5. **Factor numerator:** $$= \frac{e^x[(2x - 1) - (x^2 - x)]}{e^{2x}}$$ 6. **Simplify inside brackets:** $$(2x - 1) - (x^2 - x) = 2x - 1 - x^2 + x = -x^2 + 3x - 1$$ 7. **Rewrite derivative:** $$\frac{dy}{dx} = \frac{e^x(-x^2 + 3x - 1)}{e^{2x}}$$ 8. **Simplify fraction by canceling $e^x$:** $$= \frac{\cancel{e^x}(-x^2 + 3x - 1)}{\cancel{e^x} e^x} = \frac{-x^2 + 3x - 1}{e^x}$$ --- 1. **Problem (ii):** Find the derivative of $\tan^2(5x)$ and simplify. 2. **Rewrite function:** $$y = (\tan(5x))^2$$ 3. **Use chain rule:** $$\frac{dy}{dx} = 2\tan(5x) \cdot \frac{d}{dx}[\tan(5x)]$$ 4. **Derivative of $\tan(5x)$:** $$\frac{d}{dx}[\tan(5x)] = \sec^2(5x) \cdot 5 = 5\sec^2(5x)$$ 5. **Combine:** $$\frac{dy}{dx} = 2\tan(5x) \cdot 5\sec^2(5x) = 10\tan(5x)\sec^2(5x)$$ **Final answers:** (i) $$\frac{dy}{dx} = \frac{-x^2 + 3x - 1}{e^x}$$ (ii) $$\frac{dy}{dx} = 10\tan(5x)\sec^2(5x)$$