Find the derivatives of the following at any point of their domains :
`f(x)=(ax+b)/(cx+d)`

Answer

To find the derivative of the function \( f(x) = \frac{ax + b}{cx + d} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form of \( \frac{u}{v} \), where \( u \) and \( v \) are both functions of \( x \), then the derivative is given by:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

### Step-by-Step Solution:

1. **Identify \( u \) and \( v \)**:
- Let \( u = ax + b \)
- Let \( v = cx + d \)

2. **Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)**:
- \( \frac{du}{dx} = a \) (since the derivative of \( ax + b \) is \( a \))
- \( \frac{dv}{dx} = c \) (since the derivative of \( cx + d \) is \( c \))

3. **Apply the Quotient Rule**:
- Using the quotient rule:
\[ f'(x) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
- Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ f'(x) = \frac{(cx + d)(a) - (ax + b)(c)}{(cx + d)^2} \]

4. **Simplify the numerator**:
- Expand the numerator:
\[ f'(x) = \frac{acx + ad - acx - bc}{(cx + d)^2} \]
- Combine like terms:
\[ f'(x) = \frac{ad - bc}{(cx + d)^2} \]

5. **Final Result**:
- Therefore, the derivative of the function \( f(x) \) is:
\[ f'(x) = \frac{ad - bc}{(cx + d)^2} \]