Let `f _(1)(x) =e ^(x) and f _(n+1) (x) =e ^(f _(n)(x)))` for any `n ge 1, n in N.`Then for any fixed n, the vlaue of `(d)/(dx) f )(n)(x)` equals:

Options

A. `f _(n )(x)`
B. `f _(n)(x) f _(n-1) (x)…..f_(2)(x) f _(2)(x)`
C. `f _(n)(x)f _(n-1) (x)`
D. `f _(n)(x) f _(n-1) (x)…..f _(2)(x) f_(1)(x) e ^(x)`

Answer

To solve the problem, we need to find the derivative of the function \( f_n(x) \) defined recursively as follows:

1. \( f_1(x) = e^x \)
2. \( f_{n+1}(x) = e^{f_n(x)} \) for \( n \geq 1 \)

We want to compute \( \frac{d}{dx} f_n(x) \).

### Step-by-Step Solution:

**Step 1: Understand the recursive definition**
- We know that \( f_1(x) = e^x \).
- For \( n = 1 \), \( f_2(x) = e^{f_1(x)} = e^{e^x} \).
- For \( n = 2 \), \( f_3(x) = e^{f_2(x)} = e^{e^{e^x}} \).
- Continuing this pattern, we can see that each \( f_n(x) \) is an exponential tower of \( e \) functions.

**Step 2: Differentiate \( f_n(x) \)**
- We apply the chain rule for differentiation:
\[ \frac{d}{dx} f_n(x) = \frac{d}{dx} e^{f_{n-1}(x)} = e^{f_{n-1}(x)} \cdot \frac{d}{dx} f_{n-1}(x) \]

**Step 3: Apply the differentiation recursively**
- Now, we can express the derivative of \( f_n(x) \) in terms of \( f_{n-1}(x) \):
\[ \frac{d}{dx} f_n(x) = e^{f_{n-1}(x)} \cdot \frac{d}{dx} f_{n-1}(x) \]
- If we differentiate \( f_{n-1}(x) \) similarly, we get:
\[ \frac{d}{dx} f_{n-1}(x) = e^{f_{n-2}(x)} \cdot \frac{d}{dx} f_{n-2}(x) \]

**Step 4: Continue this process**
- Continuing this process, we can see that:
\[ \frac{d}{dx} f_n(x) = e^{f_{n-1}(x)} \cdot e^{f_{n-2}(x)} \cdots e^{f_1(x)} \cdot \frac{d}{dx} f_1(x) \]
- Since \( f_1(x) = e^x \), we have:
\[ \frac{d}{dx} f_1(x) = e^x \]

**Step 5: Substitute back**
- Thus, we can express \( \frac{d}{dx} f_n(x) \) as:
\[ \frac{d}{dx} f_n(x) = e^{f_{n-1}(x)} \cdot e^{f_{n-2}(x)} \cdots e^{f_1(x)} \cdot e^x \]

**Step 6: Final expression**
- This leads us to the final expression:
\[ \frac{d}{dx} f_n(x) = e^{f_{n-1}(x) + f_{n-2}(x) + \ldots + f_1(x) + x} \]

### Conclusion:
The value of \( \frac{d}{dx} f_n(x) \) is:
\[ \frac{d}{dx} f_n(x) = e^{f_{n-1}(x) + f_{n-2}(x) + \ldots + f_1(x) + x} \]