Differentiate `x^3 + sin 4x + e^(3x)` w.r.t `x`

Answer

To differentiate the function \( y = x^3 + \sin(4x) + e^{3x} \) with respect to \( x \), we will apply the rules of differentiation step by step.

### Step-by-Step Solution:

1. **Identify the function to differentiate**:
\[ y = x^3 + \sin(4x) + e^{3x} \]

2. **Differentiate each term separately**:
- For the first term \( x^3 \):
\[ \frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2 \]

- For the second term \( \sin(4x) \):
- We use the chain rule. The derivative of \( \sin(u) \) is \( \cos(u) \cdot \frac{du}{dx} \), where \( u = 4x \).
\[ \frac{d}{dx}(\sin(4x)) = \cos(4x) \cdot \frac{d}{dx}(4x) = 4\cos(4x) \]

- For the third term \( e^{3x} \):
- Again, we use the chain rule. The derivative of \( e^{u} \) is \( e^{u} \cdot \frac{du}{dx} \), where \( u = 3x \).
\[ \frac{d}{dx}(e^{3x}) = e^{3x} \cdot \frac{d}{dx}(3x) = 3e^{3x} \]

3. **Combine the derivatives**:
Now, we can combine all the derivatives we calculated:
\[ \frac{dy}{dx} = 3x^2 + 4\cos(4x) + 3e^{3x} \]

4. **Final answer**:
Thus, the derivative of the function \( y = x^3 + \sin(4x) + e^{3x} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = 3x^2 + 4\cos(4x) + 3e^{3x} \]