Question

If $f(1)=1$, $f^{\prime}(1)=3$, then the derivative of $f(f(f(x)))+(f(x))^{2}$ at $x=1$ is

• A. 12
• B. 19
• C. 23
• D. 33

Hint:

Logic Tip: Writing out the chain rule carefully is the only hurdle here. A 3-layer nested function $f(g(h(x)))$ will always produce exactly 3 derivative terms multiplied together: $f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$. Because $f(1)=1$, evaluating the nested parts acts as an "identity" pipeline, maintaining the value at $1$.

Answer 1

The Correct Option is D

Concept:
To find the derivative of a composite function, we use the Chain Rule recursively. For $h(x) = f(g(x))$, $h'(x) = f'(g(x)) \cdot g'(x)$. For $h(x) = [f(x)]^n$, $h'(x) = n[f(x)]^{n-1} \cdot f'(x)$.
Step 1: Differentiate the given expression with respect to x.
Let the function be $y = f(f(f(x))) + (f(x))^2$. Apply the chain rule to the first term $f(f(f(x)))$: The outer derivative is $f'(f(f(x)))$. Multiply by the derivative of the inside, which is $f'(f(x)) \cdot f'(x)$. So, the derivative of the first term is $f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x)$. Apply the power rule and chain rule to the second term $(f(x))^2$: The derivative is $2f(x) \cdot f'(x)$. Combine them to get $\frac{dy}{dx}$: $$\frac{dy}{dx} = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) + 2f(x)f'(x)$$
Step 2: Evaluate the derivative at x = 1.
Substitute $x = 1$ into the derivative expression: $$\left.\frac{dy}{dx}\right|_{x=1} = f'(f(f(1))) \cdot f'(f(1)) \cdot f'(1) + 2f(1)f'(1)$$
Step 3: Substitute the known values from the inside out.
We are given $f(1) = 1$ and $f'(1) = 3$. Work from the innermost functions outward: Since $f(1) = 1$, then $f(f(1)) = f(1) = 1$. Substituting these values: $$\left.\frac{dy}{dx}\right|_{x=1} = f'(1) \cdot f'(1) \cdot f'(1) + 2(1)(3)$$ Now substitute $f'(1) = 3$: $$\left.\frac{dy}{dx}\right|_{x=1} = (3) \cdot (3) \cdot (3) + 2(1)(3)$$ $$\left.\frac{dy}{dx}\right|_{x=1} = 27 + 6 = 33$$