c) Differentiate i. y = 4x³ (3x-1)⁴ ii. 2yx − y² = 25

$$y = 4x^3 - 3x^2 - 6x + 2$$ $$\frac{dy}{dx} = \frac{d}{dx}(4x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(6x) + \frac{d}{dx}(2)$$ $$\frac{dy}{dx} = 12x^2 - 6x - 6$$

$$12x^2 - 6x - 6 = 0$$

Divide the entire equation by 6:

$$2x^2 - x - 1 = 0$$

Factor the quadratic equation:

$$(2x+1)(x-1) = 0$$

This gives two possible values for $x$:

$$2x+1 = 0 \implies x = -\frac{1}{2}$$ $$x-1 = 0 \implies x = 1$$

For $x = -\frac{1}{2}$:

$$y = 4\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right)^2 - 6\left(-\frac{1}{2}\right) + 2$$ $$y = 4\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) + 3 + 2$$ $$y = -\frac{1}{2} - \frac{3}{4} + 5$$ $$y = -\frac{2}{4} - \frac{3}{4} + \frac{20}{4}$$ $$y = \frac{-2 - 3 + 20}{4} = \frac{15}{4}$$

So, the first turning point is $\left(-\frac{1}{2}, \frac{15}{4}\right)$.

For $x = 1$:

$$y = 4(1)^3 - 3(1)^2 - 6(1) + 2$$ $$y = 4 - 3 - 6 + 2$$ $$y = -3$$

So, the second turning point is $(1, -3)$.

Find the second derivative, $\frac{d^2y}{dx^2}$:

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(12x^2 - 6x - 6)$$ $$\frac{d^2y}{dx^2} = 24x - 6$$

Evaluate $\frac{d^2y}{dx^2}$ at each critical point:
For $x = -\frac{1}{2}$:

$$\frac{d^2y}{dx^2} = 24\left(-\frac{1}{2}\right) - 6$$ $$\frac{d^2y}{dx^2} = -12 - 6 = -18$$

Since $\frac{d^2y}{dx^2} < 0$, the point $\left(-\frac{1}{2}, \frac{15}{4}\right)$ is a local maximum.

For $x = 1$:

$$\frac{d^2y}{dx^2} = 24(1) - 6$$ $$\frac{d^2y}{dx^2} = 24 - 6 = 18$$

Since $\frac{d^2y}{dx^2} > 0$, the point $(1, -3)$ is a local minimum.

The turning points are:

• Local maximum: $\boxed{\left(-\frac{1}{2}, \frac{15}{4}\right)}$
• Local minimum: $\boxed{(1, -3)}$