Question

Let $f_{1}(s)=e^{x}, f_{2}(x)=e^{f(x)}, \ldots \ldots . f_{n+1}(x)=e^{f(x)}$ for all $n \geq 1$. The for any fixed $n, \frac{d}{d x} f_{n}(x)$ is
(A) $f_{n}(x)$
(B) $f_{n}(x) f_{n-1}(x)$
(C) $f_{n}(x) f_{n-1}(x) \ldots . f_{1}(x)$
(D) $f_{n}(x) \ldots \ldots \ldots f_{1}(x) e^{x}$

Answer 1

Ans : (C)
Hint $: \frac{d}{d x} f_{n}(x)=\frac{d}{d f_{n-1}(x)} e^{f_{n-1}(x)} \times \frac{d}{d x} f_{n-1}(x)$
$$\begin{aligned} &=f_{n}(x) \times \frac{d}{d f_{n-2}(x)} e^{f_{n-2}(x)} \times \frac{d}{d x} f_{n-2}(x) \\ &=f_{n}(x) f_{n-1}(x) f_{n-2}(x) \times \ldots \ldots \ldots . . \times f_{2}(x) \times \frac{d}{d x} f_{1}(x) \\ &=f_{n}(x) f_{n-1}(x) f_{n-2}(x) \ldots \ldots \ldots \ldots f_{1}(x) \end{aligned}$$

Answer 2

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Answer 3

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