Find the first, second, and third derivatives of y = x3 + e^x and y = sin(x).

Using the chain rule, $\frac{d}{dx}(\sin(ax)) = a\cos(ax)$:

$$\frac{dy}{dx} = 2\cos(2x)$$

Differentiate $\frac{dy}{dx} = 2\cos(2x)$. Using the chain rule, $\frac{d}{dx}(\cos(ax)) = -a\sin(ax)$:

$$\frac{d^2y}{dx^2} = 2(-2\sin(2x))$$ $$\frac{d^2y}{dx^2} = -4\sin(2x)$$

Differentiate $\frac{d^2y}{dx^2} = -4\sin(2x)$. Using the chain rule, $\frac{d}{dx}(\sin(ax)) = a\cos(ax)$:

$$\frac{d^3y}{dx^3} = -4(2\cos(2x))$$ $$\frac{d^3y}{dx^3} = \boxed{\mathbf{-8\cos(2x)}}$$

Differentiate $x^3$ using the power rule ($\frac{d}{dx}(x^n) = nx^{n-1}$) and $e^{2x}$ using the chain rule ($\frac{d}{dx}(e^{ax}) = ae^{ax}$):

$$\frac{dy}{dx} = 3x^{3-1} + 2e^{2x}$$ $$\frac{dy}{dx} = 3x^2 + 2e^{2x}$$

Differentiate $\frac{dy}{dx} = 3x^2 + 2e^{2x}$:

$$\frac{d^2y}{dx^2} = 3(2x^{2-1}) + 2(2e^{2x})$$ $$\frac{d^2y}{dx^2} = 6x + 4e^{2x}$$

Differentiate $\frac{d^2y}{dx^2} = 6x + 4e^{2x}$:

$$\frac{d^3y}{dx^3} = 6(1x^{1-1}) + 4(2e^{2x})$$ $$\frac{d^3y}{dx^3} = 6 + 8e^{2x}$$ $$\frac{d^3y}{dx^3} = \boxed{\mathbf{6 + 8e^{2x}}}$$