The question asks for the derivative of the given rational function:

$f(x) = \frac{x^2 - x + 5}{4x - 1}.$

We will solve this using the quotient rule, which states:

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

where $u(x) = x^2 - x + 5$ and $v(x) = 4x - 1$.

Step-by-step Solution:

1. Find $u'(x)$: $u(x) = x^2 - x + 5 \implies u'(x) = 2x - 1.$

2. Find $v'(x)$: $v(x) = 4x - 1 \implies v'(x) = 4.$

3. Apply the quotient rule: $f'(x) = \frac{(2x - 1)(4x - 1) - (x^2 - x + 5)(4)}{(4x - 1)^2}.$

4. Simplify the numerator:

   Expand $(2x - 1)(4x - 1)$: $(2x - 1)(4x - 1) = 8x^2 - 2x - 4x + 1 = 8x^2 - 6x + 1.$

   Expand $(x^2 - x + 5)(4)$: $(x^2 - x + 5)(4) = 4x^2 - 4x + 20.$

   Subtract the second expression from the first: $(8x^2 - 6x + 1) - (4x^2 - 4x + 20) = 8x^2 - 6x + 1 - 4x^2 + 4x - 20.$

   Combine like terms: $8x^2 - 4x^2 = 4x^2, \quad -6x + 4x = -2x, \quad 1 - 20 = -19.$

   So the numerator becomes: $4x^2 - 2x - 19.$

5. Write the final derivative: $f'(x) = \frac{4x^2 - 2x - 19}{(4x - 1)^2}.$

Final Answer:

$\boxed{\frac{4x^2 - 2x - 19}{(4x - 1)^2}}$

Would you like further clarification or details? Here are 5 related questions to explore:

1. How is the quotient rule derived?
2. What are the key differences between the product rule and quotient rule?
3. How would the solution change if $v(x)$ were a constant?
4. Can we use an alternative method like simplifying before differentiating here?
5. How do higher-order derivatives apply to rational functions?

Tip: Always simplify your numerator as much as possible when using the quotient rule to avoid mistakes in calculations.