Question

The derivative of \( y = (1 - x)(2 - x) \dots (n - x) \) at \( x = 1 \) is

• A. \( (n - 1)! \)
• B. \( n! \)
• C. \( (-1)(n - 1)! \)
• D. \( (-n)(n - 1)! \)

Hint:

In a product where one factor is zero at \(a\), the derivative at \(a\) is the derivative of that factor times the remaining product at \(a\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the derivative of a product of \( n \) terms at a specific point \( x = 1 \).
Notice that one of the terms in the product becomes zero at \( x = 1 \), which is \( (1 - x) \).
Step 2: Key Formula or Approach:
Use the product rule for differentiation. Let \( y = u(x) \cdot v(x) \), then \( y' = u'(x)v(x) + u(x)v'(x) \).
We can separate the term that becomes zero from the rest of the product.
Step 3: Detailed Explanation:
Let the given function be written as:
\[ y = (1 - x) \cdot P(x) \] where \( P(x) = (2 - x)(3 - x) \dots (n - x) \).
Now, differentiate \( y \) with respect to \( x \) using the product rule:
\[ \frac{dy}{dx} = \frac{d}{dx}(1 - x) \cdot P(x) + (1 - x) \cdot \frac{d}{dx}P(x) \] \[ \frac{dy}{dx} = (-1) \cdot P(x) + (1 - x) \cdot P'(x) \] We need to evaluate this derivative at \( x = 1 \):
\[ \left. \frac{dy}{dx} \right|_{x=1} = (-1) \cdot P(1) + (1 - 1) \cdot P'(1) \] \[ \left. \frac{dy}{dx} \right|_{x=1} = -P(1) + 0 \] Now, we calculate \( P(1) \):
\[ P(1) = (2 - 1)(3 - 1)(4 - 1) \dots (n - 1) \] \[ P(1) = (1)(2)(3) \dots (n - 1) \] This is the product of integers from 1 to \( n-1 \), which is defined as \( (n-1)! \).
Therefore,
\[ \left. \frac{dy}{dx} \right|_{x=1} = -(n - 1)! = (-1)(n - 1)! \] Step 4: Final Answer:
The derivative at \( x=1 \) is \( (-1)(n - 1)! \).