Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers) :
`(px+q)(r/x+s)`

Answer

To find the derivative of the function \( f(x) = (px + q)\left(\frac{r}{x} + s\right) \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by:

\[ (uv)' = u'v + uv' \]

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = px + q \) and \( v = \frac{r}{x} + s \).

2. **Differentiate \( u \)**:
\[ u' = \frac{d}{dx}(px + q) = p \]
(Since \( p \) is a constant and the derivative of \( q \) is 0).

3. **Differentiate \( v \)**:
To differentiate \( v = \frac{r}{x} + s \), we can rewrite \( \frac{r}{x} \) as \( r \cdot x^{-1} \):
\[ v' = \frac{d}{dx}\left(\frac{r}{x}\right) + \frac{d}{dx}(s) = -\frac{r}{x^2} + 0 = -\frac{r}{x^2} \]
(The derivative of \( s \) is 0 since it is a constant).

4. **Apply the product rule**:
Now, we apply the product rule:
\[ f'(x) = u'v + uv' \]
Substituting the values we found:
\[ f'(x) = p\left(\frac{r}{x} + s\right) + (px + q)\left(-\frac{r}{x^2}\right) \]

5. **Simplify the expression**:
Expanding the terms:
\[ f'(x) = p\left(\frac{r}{x}\right) + ps - \left(px + q\right)\left(\frac{r}{x^2}\right) \]
\[ = \frac{pr}{x} + ps - \frac{pr}{x} - \frac{qr}{x^2} \]

Notice that the \( \frac{pr}{x} \) terms cancel out:
\[ f'(x) = ps - \frac{qr}{x^2} \]

### Final Answer:
Thus, the derivative of the function is:
\[ f'(x) = ps - \frac{qr}{x^2} \]