Let `f:RtoR` be a function such that `f(x)=x^(3)+x^(2)f'(1)+xf''(2)+f'''(3)` for `x in R`
What is `f'''(10)` equal to ?

Options

A. A) 1
B. B) 5
C. C) 6
D. D) 0

Answer

To solve the problem, we need to find the value of \( f'''(10) \) given the function:

\[ f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3) \]

### Step 1: Differentiate \( f(x) \) to find \( f'(x) \)

We start by differentiating \( f(x) \):

\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2 f'(1)) + \frac{d}{dx}(x f''(2)) + \frac{d}{dx}(f'''(3)) \]

Calculating each term:

1. The derivative of \( x^3 \) is \( 3x^2 \).
2. The derivative of \( x^2 f'(1) \) (where \( f'(1) \) is a constant) is \( 2x f'(1) \).
3. The derivative of \( x f''(2) \) (where \( f''(2) \) is a constant) is \( f''(2) \).
4. The derivative of \( f'''(3) \) (where \( f'''(3) \) is a constant) is \( 0 \).

Putting it all together, we have:

\[ f'(x) = 3x^2 + 2x f'(1) + f''(2) \]

### Step 2: Differentiate \( f'(x) \) to find \( f''(x) \)

Now we differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x f'(1)) + \frac{d}{dx}(f''(2)) \]

Calculating each term:

1. The derivative of \( 3x^2 \) is \( 6x \).
2. The derivative of \( 2x f'(1) \) (where \( f'(1) \) is a constant) is \( 2 f'(1) \).
3. The derivative of \( f''(2) \) (where \( f''(2) \) is a constant) is \( 0 \).

Putting it all together, we have:

\[ f''(x) = 6x + 2f'(1) \]

### Step 3: Differentiate \( f''(x) \) to find \( f'''(x) \)

Now we differentiate \( f''(x) \):

\[ f'''(x) = \frac{d}{dx}(6x) + \frac{d}{dx}(2f'(1)) \]

Calculating each term:

1. The derivative of \( 6x \) is \( 6 \).
2. The derivative of \( 2f'(1) \) (where \( f'(1) \) is a constant) is \( 0 \).

Putting it all together, we have:

\[ f'''(x) = 6 \]

### Step 4: Evaluate \( f'''(10) \)

Since \( f'''(x) \) is a constant function, we can evaluate it at any point:

\[ f'''(10) = 6 \]

### Final Answer

Thus, the value of \( f'''(10) \) is:

\[ \boxed{6} \]