`"If "f(x)=(x-1)^(4)(x-2)^(3)(x-3)^(2)(x-4),` then the value of `f'''(1)+f''(2)+f'(3)+f'(4)` equals

Options

A. 0
B. 50
C. 324
D. 648

Explanation

To solve the problem, we need to evaluate the expression \( f'''(1) + f''(2) + f'(3) + f'(4) \) for the function \( f(x) = (x-1)^4 (x-2)^3 (x-3)^2 (x-4) \).

### Step 1: Evaluate \( f'(3) \)
Using the product rule, we differentiate \( f(x) \). However, we can directly evaluate \( f'(3) \) without fully differentiating \( f(x) \) because \( f(x) \) has a factor \( (x-3)^2 \). Thus, when we substitute \( x = 3 \):

\[ f'(3) = 0 \]

### Step 2: Evaluate \( f'(4) \)
Next, we evaluate \( f'(4) \). The term \( (x-4) \) will remain while all other factors will become zero at \( x = 4 \). Thus:

\[ f'(4) = (4-1)^4 (4-2)^3 (4-3)^2 = 3^4 \cdot 2^3 \cdot 1^2 = 81 \cdot 8 \cdot 1 = 648 \]

### Step 3: Evaluate \( f''(2) \)
For \( f''(2) \), since \( f(x) \) has a factor \( (x-2)^3 \), substituting \( x = 2 \) will yield:

\[ f''(2) = 0 \]

### Step 4: Evaluate \( f'''(1) \)
Finally, we evaluate \( f'''(1) \). The function \( f(x) \) has a factor \( (x-1)^4 \), so substituting \( x = 1 \) gives:

\[ f'''(1) = 0 \]

### Step 5: Combine the results
Now we combine all the results:

\[ f'''(1) + f''(2) + f'(3) + f'(4) = 0 + 0 + 0 + 648 = 648 \]

### Final Answer
Thus, the value of \( f'''(1) + f''(2) + f'(3) + f'(4) \) is \( \boxed{648} \).

---