Given: - \( y = f(f(f(x))) + (f(x))^2 \) - \( f'(1) = 3 \) - \( f(1) = 1 \) Differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} \left[ f(f(f(x))) \right] + \frac{d}{dx} \left[ (f(x))^2 \right] \] For the first term, \( \frac{d}{dx} [f(f(f(x)))] \), using the chain rule repeatedly: Let \( u = f(x) \), then \( \frac{du}{dx} = f'(x) \). Let \( v = f(u) = f(f(x)) \), then \( \frac{dv}{dx} = f'(f(x)) \cdot f'(x) \). Therefore, \[ \frac{d}{dx} [f(f(f(x)))] = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) \] For the second term, \( \frac{d}{dx} [(f(x))^2] \), using the chain rule: \[ \frac{d}{dx} [(f(x))^2] = 2 f(x) f'(x) \] Evaluate at \( x = 1 \): - Since \( f(1) = 1 \), then \( f(f(1)) = f(1) = 1 \), and \( f(f(f(1))) = f(1) = 1 \). This implies that for the derivative of the first term, all inner function evaluations at \( x=1 \) result in 1. - We are given \( f'(1) = 3 \). Thus, the derivative of the first term at \( x=1 \) is: \[ f'(f(f(1))) \cdot f'(f(1)) \cdot f'(1) = f'(1) \cdot f'(1) \cdot f'(1) = 3 \cdot 3 \cdot 3 = 27 \] The derivative of the second term at \( x=1 \) is: \[ 2 \cdot f(1) \cdot f'(1) = 2 \cdot 1 \cdot 3 = 6 \] Combining both terms: \[ \frac{dy}{dx} \Big|_{x=1} = 27 + 6 = 33 \] The final correct answer is: \[ \boxed{33} \] Correct Answer: (None from given options, actual answer is 33)