Question

Given \( f'(1) = 3 \), \( f(1) = 1 \), and
\[ y = f\left(f(f(x))\right) + \left(f(x)\right)^2, \] then find \( \frac{dy}{dx} \) at \( x = 1 \).

• A. \( 9 \)
• B. \( 12 \)
• C. \( 15 \)
• D. \( 18 \)

Hint:

When differentiating composite functions like \( f(f(f(x))) \), apply the chain rule multiple times step by step and plug in given values carefully.

Answer 1

The Correct Option is C

We are given: - \( y = f(f(f(x))) + (f(x))^2 \) - \( f'(1) = 3 \) - \( f(1) = 1 \) Let us differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} \left[ f(f(f(x))) \right] + \frac{d}{dx} \left[ (f(x))^2 \right] \] Term 1: \( f(f(f(x))) \) Let \( u = f(x) \Rightarrow \frac{du}{dx} = f'(x) \) Let \( v = f(u) = f(f(x)) \Rightarrow \frac{dv}{dx} = f'(f(x)) \cdot f'(x) \) Then, \[ \frac{d}{dx} [f(f(f(x)))] = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) \] Term 2: \( (f(x))^2 \) \[ \frac{d}{dx} [(f(x))^2] = 2 f(x) f'(x) \] Now plug in \( x = 1 \): - \( f(1) = 1 \Rightarrow f(f(1)) = f(1) = 1 \Rightarrow f(f(f(1))) = f(1) = 1 \) - \( f'(1) = 3 \) So, \[ \frac{dy}{dx} = f'(1) \cdot f'(1) \cdot f'(1) + 2 \cdot f(1) \cdot f'(1) = 3 \cdot 3 \cdot 3 + 2 \cdot 1 \cdot 3 = 27 + 6 = 33 \] Correction! Let’s double-check. Since: \[ f(f(f(x))) \text{ derivative } = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) \] At \( x = 1 \): - \( f(1) = 1 \Rightarrow f(f(1)) = f(1) = 1 \Rightarrow f(f(f(1))) = f(1) = 1 \) - So all inner function values are 1 Thus, \[ f'(f(f(1))) = f'(1),\quad f'(f(1)) = f'(1),\quad f'(1) = 3 \Rightarrow f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) = 3 \cdot 3 \cdot 3 = 27 \] Now: \[ \frac{dy}{dx} = 27 + 2 \cdot f(1) \cdot f'(1) = 27 + 2 \cdot 1 \cdot 3 = 27 + 6 = \boxed{33} \] So the final correct answer is: \[ \boxed{33} \] Correct Answer: (None from given options, actual answer is 33)