If `f(x) = (x - 1)(x - 2)(x - 3)(x - 4)(x - 5)`, then the value of f' (5) is equal to

Options

A. 0
B. 120
C. 24
D. 5

Answer

To find the value of \( f'(5) \) for the function \( f(x) = (x - 1)(x - 2)(x - 3)(x - 4)(x - 5) \), we will use the product rule of differentiation.

### Step-by-step Solution:

1. **Identify the Function**:
\[ f(x) = (x - 1)(x - 2)(x - 3)(x - 4)(x - 5) \]

2. **Differentiate Using the Product Rule**:
The product rule states that if you have a product of functions, the derivative is given by:
\[ (uv)' = u'v + uv' \]
For our case, we have five functions multiplied together. We will differentiate each function one at a time while keeping the others constant.

3. **Apply the Product Rule**:
Let's denote:
- \( g_1(x) = (x - 1) \)
- \( g_2(x) = (x - 2) \)
- \( g_3(x) = (x - 3) \)
- \( g_4(x) = (x - 4) \)
- \( g_5(x) = (x - 5) \)

The derivative \( f'(x) \) can be expressed as:
\[ f'(x) = g_1'(x)g_2(x)g_3(x)g_4(x)g_5(x) + g_1(x)g_2'(x)g_3(x)g_4(x)g_5(x) + g_1(x)g_2(x)g_3'(x)g_4(x)g_5(x) + g_1(x)g_2(x)g_3(x)g_4'(x)g_5(x) + g_1(x)g_2(x)g_3(x)g_4(x)g_5'(x) \]

Each derivative \( g_i'(x) = 1 \) since they are linear functions.

4. **Evaluate at \( x = 5 \)**:
When we substitute \( x = 5 \):
- \( g_1(5) = 4 \)
- \( g_2(5) = 3 \)
- \( g_3(5) = 2 \)
- \( g_4(5) = 1 \)
- \( g_5(5) = 0 \)

Notice that \( g_5(5) = 0 \) will cause the last term to vanish, and similarly for any term where \( g_i(5) \) is zero.

Therefore, we only need to consider the terms where \( g_i(5) \) is not zero:
\[ f'(5) = g_1'(5)g_2(5)g_3(5)g_4(5)g_5(5) + g_1(5)g_2'(5)g_3(5)g_4(5)g_5(5) + g_1(5)g_2(5)g_3'(5)g_4(5)g_5(5) + g_1(5)g_2(5)g_3(5)g_4'(5)g_5(5) \]
Simplifying this gives:
\[ f'(5) = 1 \cdot 3 \cdot 2 \cdot 1 \cdot 0 + 4 \cdot 1 \cdot 2 \cdot 1 \cdot 0 + 4 \cdot 3 \cdot 1 \cdot 1 \cdot 0 + 4 \cdot 3 \cdot 2 \cdot 1 \cdot 0 \]
All terms vanish except for the last term:
\[ f'(5) = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \]

5. **Final Answer**:
\[ f'(5) = 24 \]