If `f(1) = 1, f'(1) = 3`, then the value of derivative of `f(f(f(x))` + `(f(x))^(2)` at `x = 1` is

Options

A. 9
B. 33
C. 12
D. 20

Answer

To solve the problem, we need to find the derivative of the function \( g(x) = f(f(f(x))) + (f(x))^2 \) at \( x = 1 \). Given that \( f(1) = 1 \) and \( f'(1) = 3 \), we will use the chain rule and product rule for differentiation.

### Step-by-Step Solution:

1. **Define the function**:
\[ g(x) = f(f(f(x))) + (f(x))^2 \]

2. **Differentiate \( g(x) \)**:
To find \( g'(x) \), we apply the chain rule and product rule:
\[ g'(x) = \frac{d}{dx}[f(f(f(x)))] + \frac{d}{dx}[(f(x))^2] \]

- For the first term \( f(f(f(x))) \):
Using the chain rule:
\[ \frac{d}{dx}[f(f(f(x)))] = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) \]

- For the second term \( (f(x))^2 \):
Using the product rule:
\[ \frac{d}{dx}[(f(x))^2] = 2f(x) \cdot f'(x) \]

Therefore, we have:
\[ g'(x) = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) + 2f(x) \cdot f'(x) \]

3. **Evaluate \( g'(1) \)**:
Now we need to substitute \( x = 1 \):
\[ g'(1) = f'(f(f(1))) \cdot f'(f(1)) \cdot f'(1) + 2f(1) \cdot f'(1) \]

- We know \( f(1) = 1 \), so:
\[ f(f(1)) = f(1) = 1 \]
\[ f(f(f(1))) = f(f(1)) = f(1) = 1 \]

- Now substitute the known values:
\[ g'(1) = f'(1) \cdot f'(1) \cdot f'(1) + 2f(1) \cdot f'(1) \]
\[ g'(1) = f'(1)^3 + 2f(1) \cdot f'(1) \]

- Using \( f'(1) = 3 \) and \( f(1) = 1 \):
\[ g'(1) = 3^3 + 2 \cdot 1 \cdot 3 \]
\[ g'(1) = 27 + 6 = 33 \]

### Final Answer:
The value of the derivative \( g'(1) \) is \( 33 \).