1. **Problem (a):** Find the gradient of the curve $y = 3x^4 - 2x^2 + 5x - 2$ at points $(0, -2)$ and $(1, 4)$. 2. **Formula:** The gradient of a curve at any point is given by the derivative $\frac{dy}{dx}$. 3. **Step:** Differentiate $y$ with respect to $x$: $$\frac{dy}{dx} = \frac{d}{dx}(3x^4) - \frac{d}{dx}(2x^2) + \frac{d}{dx}(5x) - \frac{d}{dx}(2) = 12x^3 - 4x + 5$$ 4. **Evaluate gradient at $x=0$:** $$\frac{dy}{dx}\bigg|_{x=0} = 12(0)^3 - 4(0) + 5 = 5$$ 5. **Evaluate gradient at $x=1$:** $$\frac{dy}{dx}\bigg|_{x=1} = 12(1)^3 - 4(1) + 5 = 12 - 4 + 5 = 13$$ --- 6. **Problem (b):** Find coordinates on $y = 3x^2 - 7x + 2$ where gradient is $-1$. 7. **Step:** Differentiate $y$: $$\frac{dy}{dx} = 6x - 7$$ 8. **Set gradient equal to $-1$ and solve for $x$:** $$6x - 7 = -1$$ $$6x = 6$$ $$x = 1$$ 9. **Find $y$ at $x=1$:** $$y = 3(1)^2 - 7(1) + 2 = 3 - 7 + 2 = -2$$ 10. **Coordinates:** $(1, -2)$ --- 11. **Problem (c):** Find differential coefficient of $y = 3x^2 \sin 2x$. 12. **Step:** Use product rule $\frac{d}{dx}(uv) = u'v + uv'$ where $u=3x^2$, $v=\sin 2x$. 13. **Calculate derivatives:** $$u' = 6x$$ $$v' = 2 \cos 2x$$ 14. **Apply product rule:** $$\frac{dy}{dx} = 6x \sin 2x + 3x^2 (2 \cos 2x) = 6x \sin 2x + 6x^2 \cos 2x$$