Find the derivatives of the following (9-10) at any point of their domains :
`H(x)=((x-1)(x-2))/((x-3)(x-4)), x ne 3, 4`

Answer

To find the derivative of the function \( H(x) = \frac{(x-1)(x-2)}{(x-3)(x-4)} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then its derivative is given by:

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

where \( u \) is the numerator and \( v \) is the denominator.

### Step-by-step Solution:

1. **Identify \( u \) and \( v \)**:
- Let \( u = (x-1)(x-2) \)
- Let \( v = (x-3)(x-4) \)

2. **Differentiate \( u \)**:
- First, expand \( u \):
\[ u = (x-1)(x-2) = x^2 - 3x + 2 \]
- Now differentiate \( u \):
\[ u' = \frac{d}{dx}(x^2 - 3x + 2) = 2x - 3 \]

3. **Differentiate \( v \)**:
- First, expand \( v \):
\[ v = (x-3)(x-4) = x^2 - 7x + 12 \]
- Now differentiate \( v \):
\[ v' = \frac{d}{dx}(x^2 - 7x + 12) = 2x - 7 \]

4. **Apply the Quotient Rule**:
- Now substitute \( u, u', v, \) and \( v' \) into the quotient rule formula:
\[ H'(x) = \frac{(2x - 3)(x^2 - 7x + 12) - (x^2 - 3x + 2)(2x - 7)}{(x^2 - 7x + 12)^2} \]

5. **Simplify the numerator**:
- Expand both products in the numerator:
- First term:
\[ (2x - 3)(x^2 - 7x + 12) = 2x^3 - 14x^2 + 24x - 3x^2 + 21x - 36 = 2x^3 - 17x^2 + 45x - 36 \]
- Second term:
\[ (x^2 - 3x + 2)(2x - 7) = 2x^3 - 7x^2 - 6x^2 + 21x + 4x - 14 = 2x^3 - 13x^2 + 25x - 14 \]
- Combine the results:
\[ H'(x) = \frac{(2x^3 - 17x^2 + 45x - 36) - (2x^3 - 13x^2 + 25x - 14)}{(x^2 - 7x + 12)^2} \]
- Simplifying the numerator:
\[ H'(x) = \frac{-4x^2 + 20x - 22}{(x^2 - 7x + 12)^2} \]

6. **Final Result**:
- The derivative of \( H(x) \) is:
\[ H'(x) = \frac{-4x^2 + 20x - 22}{(x^2 - 7x + 12)^2} \]