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

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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

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kritika · Answer 1 · 2021-12-27T06:37:51+0000

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

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

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