1. **Differentiate the function** $y = x^3 - 3x^2 + 5x - 3$ with respect to $x$. 2. **Recall the power rule for differentiation:** If $y = x^n$, then $\frac{dy}{dx} = nx^{n-1}$. 3. **Apply the power rule to each term:** $$\frac{dy}{dx} = 3x^2 - 6x + 5$$ --- 1. **Differentiate using the Quotient Rule:** Given $y = \frac{x^2}{x - 4}$. 2. **Quotient rule formula:** $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$$ where $u = x^2$, $u' = 2x$, $v = x - 4$, and $v' = 1$. 3. **Substitute into the formula:** $$\frac{dy}{dx} = \frac{(2x)(x - 4) - (x^2)(1)}{(x - 4)^2}$$ 4. **Simplify the numerator:** $$= \frac{2x^2 - 8x - x^2}{(x - 4)^2} = \frac{x^2 - 8x}{(x - 4)^2}$$ --- 1. **Find turning points of** $y = 2x^3 - 3x^2 - 12x + 7$. 2. **Find the first derivative:** $$\frac{dy}{dx} = 6x^2 - 6x - 12$$ 3. **Set derivative equal to zero to find critical points:** $$6x^2 - 6x - 12 = 0$$ 4. **Divide both sides by 6:** $$\cancel{6}x^2 - \cancel{6}x - \cancel{12} = 0 \Rightarrow x^2 - x - 2 = 0$$ 5. **Factor the quadratic:** $$ (x - 2)(x + 1) = 0 $$ 6. **Solve for $x$:** $$x = 2, \quad x = -1$$ 7. **These are the $x$-coordinates of the turning points.** --- **Final answers:** - (a) $\frac{dy}{dx} = 3x^2 - 6x + 5$ - (b) $\frac{dy}{dx} = \frac{x^2 - 8x}{(x - 4)^2}$ - (2) Turning points at $x = 2$ and $x = -1$