y = sin⁴(2x⁵ + pi/6) and y = sin(e^(3x) + 4)

Let $u = \sin(2x^5 + \frac{\pi}{6})$. Then $y = u^4$.

$$\frac{dy}{du} = 4u^3$$

Let $v = 2x^5 + \frac{\pi}{6}$. Then $u = \sin(v)$.

$$\frac{du}{dv} = \cos(v)$$ $$\frac{dv}{dx} = \frac{d}{dx}(2x^5 + \frac{\pi}{6}) = 10x^4$$

$$\frac{dy}{dx} = 4u^3 \cdot \cos(v) \cdot 10x^4$$

Substitute back $u = \sin(2x^5 + \frac{\pi}{6})$ and $v = 2x^5 + \frac{\pi}{6}$.

$$\frac{dy}{dx} = 4(\sin(2x^5 + \frac{\pi}{6}))^3 \cdot \cos(2x^5 + \frac{\pi}{6}) \cdot 10x^4$$

$$\frac{dy}{dx} = 40x^4 \sin^3(2x^5 + \frac{\pi}{6}) \cos(2x^5 + \frac{\pi}{6})$$

The final answer is $\boxed{40x^4 \sin^3(2x^5 + \frac{\pi}{6}) \cos(2x^5 + \frac{\pi}{6})}$

Let $u = e^{3x} + 4$. Then $y = \sin(u)$.

$$\frac{dy}{du} = \cos(u)$$

$$\frac{du}{dx} = \frac{d}{dx}(e^{3x} + 4)$$ $$\frac{du}{dx} = 3e^{3x} + 0 = 3e^{3x}$$

$$\frac{dy}{dx} = \cos(u) \cdot 3e^{3x}$$

Substitute back $u = e^{3x} + 4$.

$$\frac{dy}{dx} = \cos(e^{3x} + 4) \cdot 3e^{3x}$$

$$\frac{dy}{dx} = 3e^{3x} \cos(e^{3x} + 4)$$

The final answer is $\boxed{3e^{3x} \cos(e^{3x} + 4)}$