Find the derivative of the following functions using the chain rule and product rule: i) y = (2x3 − 3x + 5)4, ii) y = (3x3 + 7)3, iii) y = (x4 − 4)4.

Let $f(x) = 2x^3 - 3x + 5$ and $n=4$.

$$\frac{dy}{dx} = 4(2x^3 - 3x + 5)^{4-1} \cdot \frac{d}{dx}(2x^3 - 3x + 5)$$

$$\frac{dy}{dx} = 4(2x^3 - 3x + 5)^3 (6x^2 - 3)$$ $$\frac{dy}{dx} = 12(2x^2 - 1)(2x^3 - 3x + 5)^3$$

The derivative is $\boxed{\mathbf{12(2x^2 - 1)(2x^3 - 3x + 5)^3}}$.

Let $f(x) = 3x^3 + 7$ and $n=3$.

$$\frac{dy}{dx} = 3(3x^3 + 7)^{3-1} \cdot \frac{d}{dx}(3x^3 + 7)$$

$$\frac{dy}{dx} = 3(3x^3 + 7)^2 (9x^2)$$ $$\frac{dy}{dx} = 27x^2(3x^3 + 7)^2$$

The derivative is $\boxed{\mathbf{27x^2(3x^3 + 7)^2}}$.

Let $f(x) = x^4 - 4$ and $n=4$.

$$\frac{dy}{dx} = 4(x^4 - 4)^{4-1} \cdot \frac{d}{dx}(x^4 - 4)$$

$$\frac{dy}{dx} = 4(x^4 - 4)^3 (4x^3)$$ $$\frac{dy}{dx} = 16x^3(x^4 - 4)^3$$

The derivative is $\boxed{\mathbf{16x^3(x^4 - 4)^3}}$.

Let $u = (2x-1)^2$ and $v = (3x^2+1)^2$.

$$\frac{dy}{dx} = u'v + uv'$$

$$\frac{dy}{dx} = 4(2x-1)(3x^2+1)(9x^2-3x+1)$$

The derivative is $\boxed{\mathbf{4(2x-1)(3x^2+1)(9x^2-3x+1)}}$.