Differentiate the following functions w.r.t. x:
`sec (px + q)`

Answer

To differentiate the function \( f(x) = \sec(px + q) \) with respect to \( x \), we will use the chain rule. The chain rule states that if you have a composite function, the derivative can be found by differentiating the outer function and then multiplying it by the derivative of the inner function.

### Step-by-step Solution:

1. **Identify the outer and inner functions**:
- Outer function: \( u = \sec(v) \) where \( v = px + q \)
- Inner function: \( v = px + q \)

2. **Differentiate the outer function**:
- The derivative of \( \sec(v) \) with respect to \( v \) is \( \sec(v) \tan(v) \).
- Therefore, \( \frac{du}{dv} = \sec(px + q) \tan(px + q) \).

3. **Differentiate the inner function**:
- The derivative of \( v = px + q \) with respect to \( x \) is \( p \).
- Therefore, \( \frac{dv}{dx} = p \).

4. **Apply the chain rule**:
- According to the chain rule, \( \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} \).
- Substituting the derivatives we found:
\[ \frac{du}{dx} = \sec(px + q) \tan(px + q) \cdot p \]

5. **Final result**:
- Thus, the derivative of \( f(x) = \sec(px + q) \) with respect to \( x \) is:
\[ f'(x) = p \sec(px + q) \tan(px + q) \]

### Final Answer:
\[ \frac{d}{dx} \sec(px + q) = p \sec(px + q) \tan(px + q) \]