If $f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)$, then the value of $f'(5)$ is equal to
Step-by-Step Solution
Key Concept: When evaluating derivatives at a point where a factor becomes zero, many terms in the expansion vanish, simplifying the calculation
$f(x) = (x-2)(x-3)(x-4)(x-5) + \ldots$. Notice that at $x = 5$, most terms vanish because they contain the factor $(x-5)$. Only the term $(x-1)(x-2)(x-3)(x-4)$ contributes when $x = 5$, giving $f(5) = 4 \cdot 3 \cdot 2 \cdot 1 = 24$. Using the product rule for differentiation and evaluating at $x = 5$: each product containing $(x-5)$ vanishes, and we're left with derivatives of $(x-1)(x-2)(x-3)(x-4)$. By the product rule and setting $x = 5$, we get $f'(5) = 24$.
Correct Answer: 24