Find the derivative of the following functions:
`(ax^(2)+sinx)(p+qcosx)`

Answer

To find the derivative of the function \( (ax^2 + \sin x)(p + q \cos x) \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution:

1. **Identify the Functions**:
Let \( u = ax^2 + \sin x \) and \( v = p + q \cos x \).

2. **Differentiate \( u \)**:
\[ \frac{du}{dx} = \frac{d}{dx}(ax^2) + \frac{d}{dx}(\sin x) \]
\[ \frac{du}{dx} = 2ax + \cos x \]

3. **Differentiate \( v \)**:
\[ \frac{dv}{dx} = \frac{d}{dx}(p) + \frac{d}{dx}(q \cos x) \]
Since \( p \) is a constant, \( \frac{d}{dx}(p) = 0 \). For \( q \cos x \):
\[ \frac{d}{dx}(q \cos x) = -q \sin x \]
Thus,
\[ \frac{dv}{dx} = 0 - q \sin x = -q \sin x \]

4. **Apply the Product Rule**:
Using the product rule:
\[ \frac{d}{dx}((ax^2 + \sin x)(p + q \cos x)) = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values we found:
\[ = (ax^2 + \sin x)(-q \sin x) + (p + q \cos x)(2ax + \cos x) \]

5. **Expand the Expression**:
\[ = -q \sin x (ax^2 + \sin x) + (p + q \cos x)(2ax + \cos x) \]
Expanding both parts:
\[ = -q \sin x \cdot ax^2 - q \sin^2 x + p(2ax + \cos x) + q \cos x(2ax + \cos x) \]

6. **Combine Like Terms**:
\[ = -q \sin x \cdot ax^2 - q \sin^2 x + 2apx + p \cos x + 2aqx \cos x + q \cos^2 x \]

7. **Final Expression**:
The final expression for the derivative is:
\[ = -q \sin x \cdot ax^2 + 2apx + (2aqx + p) \cos x - q \sin^2 x + q \cos^2 x \]