Question 3: Find the derivative of the function y = (3x+2) + e2x.

Let $u = 3x+2$. Then the term becomes $\sin(u)$.
The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$. So, $f'(u) = \cos(u)$.
The derivative of $u = 3x+2$ with respect to $x$ is $\frac{du}{dx} = 3$.
Applying the chain rule:

$$\frac{d}{dx}(\sin(3x+2)) = \cos(3x+2) \cdot \frac{d}{dx}(3x+2)$$ $$\frac{d}{dx}(\sin(3x+2)) = \cos(3x+2) \cdot 3$$ $$\frac{d}{dx}(\sin(3x+2)) = 3\cos(3x+2)$$

Let $v = 2x$. Then the term becomes $e^v$.
The derivative of $e^v$ with respect to $v$ is $e^v$. So, $f'(v) = e^v$.
The derivative of $v = 2x$ with respect to $x$ is $\frac{dv}{dx} = 2$.
Applying the chain rule:

$$\frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x)$$ $$\frac{d}{dx}(e^{2x}) = e^{2x} \cdot 2$$ $$\frac{d}{dx}(e^{2x}) = 2e^{2x}$$

Since $y = \sin(3x+2) + e^{2x}$, the total derivative $\frac{dy}{dx}$ is the sum of the derivatives found in Step 1 and Step 2.

$$\frac{dy}{dx} = \frac{d}{dx}(\sin(3x+2)) + \frac{d}{dx}(e^{2x})$$ $$\frac{dy}{dx} = 3\cos(3x+2) + 2e^{2x}$$

The derivative is $\mathbf{3\cos(3x+2) + 2e^{2x}}$.