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

The given problem involves a recursive function defined as \(f_n(x) = e^{f_{n-1}(x)}\) for \(n \in \mathbb{N}\) and \(f_0(x) = x\). We need to find the derivative \(\frac{d}{dx}\{f_n(x)\}\)

1. Start with \(f_0(x) = x\).
2. For \(n = 1\), \(f_1(x) = e^{f_0(x)} = e^x\). The derivative is: \(\frac{d}{dx} f_1(x) = \frac{d}{dx} e^x = e^x = f_1(x)\).
3. For \(n = 2\), \(f_2(x) = e^{f_1(x)} = e^{e^x}\). The derivative is: \(\frac{d}{dx} f_2(x) = \frac{d}{dx}(e^{e^x}) = e^{e^x} \cdot e^x = f_2(x) \cdot f_1(x)\).
4. Let's generalize for any \(n\): \(f_n(x) = e^{f_{n-1}(x)}\). Using the chain rule: \(\frac{d}{dx} f_n(x) = e^{f_{n-1}(x)} \cdot \frac{d}{dx} f_{n-1}(x) = f_n(x) \cdot \frac{d}{dx} f_{n-1}(x)\).
5. This recursive derivative leads us to a product of derivatives: \(\frac{d}{dx} f_n(x) = f_n(x) \cdot f_{n-1}(x) \cdot \ldots \cdot f_2(x) \cdot f_1(x)\).

Therefore, the derivative \(\frac{d}{dx}\{f_n(x)\}\) is equal to \(f_n(x) \cdot f_{n-1}(x) \cdot \ldots \cdot f_2(x) \cdot f_1(x)\), matching the third option.