Question

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

• A. 10
• B. 35
• C. 33
• D. 12

Answer 1

The Correct Option is C

Given:
\(f(1) = 1\)
\(f'(1) = 3\)
Let's break down the expression step by step.

First, let's consider the term \(f(f(f(x)))\). Using the chain rule, the derivative of \(f(f(f(x)))\) with respect to x is:
\(\frac{d}{dx} \left( f(f(f(x))) \right) = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x)\)

Now, let's evaluate this at x = 1:
\(\left. \frac{d}{dx} \left( f(f(f(x))) \right) \right|_{x=1} = f'(f(f(1))) \cdot f'(f(1)) \cdot f'(1)\)

Since \(f(1) = 1\), we have:
\(\left. \frac{d}{dx} \left( f(f(f(x))) \right) \right|_{x=1} = f'(f(f(1))) \cdot f'(f(1)) \cdot 3\)

Next, let's consider the term \((f(x))^2\). The derivative of \((f(x))^2\) with respect to x is:
\(\frac{d}{dx} \left( (f(x))^2 \right) = 2 \cdot f(x) \cdot f'(x)\)

Now, let's evaluate this at x = 1:
\(\left. \frac{d}{dx} \left( (f(x))^2 \right) \right|_{x=1} = 2 \cdot f(1) \cdot f'(1)\)

Since \(f(1) = 1\) and \(f'(1) = 3\), we have:
\(\left. \frac{d}{dx} \left( (f(x))^2 \right) \right|_{x=1} = 2 \cdot 1 \cdot 3 = 6\)

Finally, to find the derivative of the entire expression, we add the derivatives of the two terms:

\(\left. \frac{d}{dx} \left( f(f(f(x))) + (f(x))^2 \right) \right|_{x=1} = \left. \frac{d}{dx} \left( f(f(f(x))) \right) \right|_{x=1} + \left. \frac{d}{dx} \left( (f(x))^2 \right) \right|_{x=1}\)

= \(f'(f(f(1))) \cdot f'(f(1)) \cdot 3 + 6\)

Since \(f(f(1)) = f(1) = 1,\) we have:

\(\left. \frac{d}{dx} \left( f(f(f(x))) + (f(x))^2 \right) \right|_{x=1} = f'(1) \cdot f'(1) \cdot 3 + 6\)
\(=3 \times 3 \times 3 + 6\)
\(= 27 + 6\)
\(= 33\)
Therefore, the derivative of \(f(f(f(1))) + (f(1))^2\) is 33, which corresponds to option (C) 33.