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

We are given the following information:

\[ f(1) = 1 \quad \text{and} \quad f'(1) = 3 \]

Let's break down the expression step by step.

First, 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, 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.

Answer 2

Let \[ y = f(f(f(x))) + (f(x))^2 \] Differentiate using chain rule: \[ \frac{dy}{dx} = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) + 2f(x)f'(x) \] Given: \[ f(1) = 1,\quad f'(1) = 3 \] Now evaluate each part: - \(f(x) = 1\) at \(x=1\) - \(f(f(x)) = f(1) = 1\) - \(f(f(f(x))) = f(1) = 1\) So, \[ f'(f(f(x))) = f'(1) = 3, \quad f'(f(x)) = f'(1) = 3, \quad f'(x) = f'(1) = 3 \] Now plug in: \[ \frac{dy}{dx} = 3 \cdot 3 \cdot 3 + 2 \cdot 1 \cdot 3 = 27 + 6 = \mathbf{33} \] Therefore, the derivative at \(x = 1\) is 33 

Answer 3

Let \(g(x) = f(f(f(x))) + (f(x))^2\). We want to find g'(1).

First, we need to find g'(x) using the chain rule and the power rule:

\(g'(x) = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) + 2f(x) \cdot f'(x)\)

Now, evaluate g'(1) using the given values f(1) = 1 and f'(1) = 3:

\(g'(1) = f'(f(f(1))) \cdot f'(f(1)) \cdot f'(1) + 2f(1) \cdot f'(1)\)

\(g'(1) = f'(f(1)) \cdot f'(1) \cdot f'(1) + 2(1) \cdot (3)\)

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

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

\(g'(1) = 27 + 6\)

\(g'(1) = 33\)

Therefore, the derivative of \(f(f(f(x))) + (f(x))^2\) at x=1 is 33.

Answer: 33