Question

If \(f_n(x) = e^{f_{n-1}(x)}\) for all \(n \in \mathbb{N}\) and \(f_0(x) = x\) then \(\frac{d}{dx}\{f_n(x)\}\) is equal to

• A. \(f_n(x) f_{n-1}(x) dx\)
• B. \(f_n(x) \frac{d}{dx}\{f_{n+1}(x)\}\)
• C. \(f_n(x) \cdot f_{n-1}(x) \cdot \ldots \cdot f_2(x) \cdot f_1(x)\)
• D. None of the above

Hint:

Derivative of \(e^{g(x)} = e^{g(x)} \cdot g'(x)\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Use chain rule repeatedly.
Step 2: Detailed Explanation:
\(f_1(x) = e^x\), \(f_2(x) = e^{e^x}\), etc.
\(\frac{d}{dx}f_n(x) = f_n(x) \cdot f_{n-1}(x) \cdot \ldots \cdot f_1(x)\)
Because derivative of \(e^{f_{n-1}} = e^{f_{n-1}} \cdot f'_{n-1}\)
Step 3: Final Answer:
Product of all \(f_k(x)\) from \(k = 1\) to \(n\).