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

Answer: (C)

33

Answer 2

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.