Step 1: Understanding the Concept:
We need to find the derivative of a composite function evaluated at a specific point.
This requires the repeated application of the chain rule.
Step 2: Key Formula or Approach:
Chain rule for $h(x) = f(g(x))$ is $h'(x) = f'(g(x)) \cdot g'(x)$.
For a triple composition $f(f(f(x)))$, the derivative is $f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x)$.
Power rule with chain rule for $(f(x))^n$ is $n(f(x))^{n-1} \cdot f'(x)$.
Step 3: Detailed Explanation:
Let the function be $g(x) = f(f(f(x))) + (f(x))^2$.
We need to find $g'(1)$.
First, find the general derivative $g'(x)$ using the chain rule:
\[ g'(x) = \frac{d}{dx}[f(f(f(x)))] + \frac{d}{dx}[(f(x))^2] \]
\[ g'(x) = f'(f(f(x))) \cdot \frac{d}{dx}[f(f(x))] + 2(f(x))^1 \cdot f'(x) \]
\[ g'(x) = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) + 2f(x)f'(x) \]
Now, evaluate this derivative at $x = 1$:
\[ g'(1) = f'(f(f(1))) \cdot f'(f(1)) \cdot f'(1) + 2f(1)f'(1) \]
We are given the values: $f(1) = 1$ and $f'(1) = 3$.
Let's evaluate the nested function values step-by-step:
Inner-most: $f(1) = 1$
Middle: $f(f(1)) = f(1) = 1$
Outer: $f(f(f(1))) = f(1) = 1$
Now substitute these back into the expression for $g'(1)$:
\[ g'(1) = f'(1) \cdot f'(1) \cdot f'(1) + 2 \cdot f(1) \cdot f'(1) \]
Substitute the numerical values $f(1)=1$ and $f'(1)=3$:
\[ g'(1) = (3) \cdot (3) \cdot (3) + 2 \cdot (1) \cdot (3) \]
\[ g'(1) = 27 + 6 \]
\[ g'(1) = 33 \]
Step 4: Final Answer:
The value of the derivative is $33$.