Differentiate each of the following with respect to x in following question: `(ax^(2)+cotx)(p+qcosx)`

Answer

To differentiate the expression \((ax^2 + \cot x)(p + q \cos x)\) with respect to \(x\), 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:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution:

1. **Identify the functions**:
Let:
\[ u = ax^2 + \cot x \]
\[ v = p + q \cos x \]

2. **Differentiate \(v\)**:
To find \(\frac{dv}{dx}\):
- The derivative of \(p\) (a constant) is \(0\).
- The derivative of \(q \cos x\) is \(-q \sin x\) (using the derivative of \(\cos x\)).

Therefore,
\[ \frac{dv}{dx} = 0 - q \sin x = -q \sin x \]

3. **Differentiate \(u\)**:
To find \(\frac{du}{dx}\):
- The derivative of \(ax^2\) is \(2ax\) (using the power rule).
- The derivative of \(\cot x\) is \(-\csc^2 x\).

Therefore,
\[ \frac{du}{dx} = 2ax - \csc^2 x \]

4. **Apply the product rule**:
Now, using the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substitute \(u\), \(v\), \(\frac{dv}{dx}\), and \(\frac{du}{dx}\):
\[ \frac{dy}{dx} = (ax^2 + \cot x)(-q \sin x) + (p + q \cos x)(2ax - \csc^2 x) \]

5. **Simplify the expression**:
Distributing the terms:
\[ \frac{dy}{dx} = -q \sin x (ax^2 + \cot x) + (p + q \cos x)(2ax - \csc^2 x) \]

### Final Answer:
The derivative of the given expression is:
\[ \frac{dy}{dx} = -q \sin x (ax^2 + \cot x) + (p + q \cos x)(2ax - \csc^2 x) \]