Let f_1(x) = e^x, f_2(x) = e^{f_1(x)}, ldots, | vedclass

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(C) Given $f_1(x) = e^x$ and $f_{n+1}(x) = e^{f_n(x)}$.Taking the natural logarithm on both sides of $f_n(x) = e^{f_{n-1}(x)}$, we get $\ln(f_n(x)) =

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$f_n(x) f_{n-1}(x) \ldots f_1(x)$

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(C) Given $f_1(x) = e^x$ and $f_{n+1}(x) = e^{f_n(x)}$.Taking the natural logarithm on both sides of $f_n(x) = e^{f_{n-1}(x)}$, we get $\ln(f_n(x)) = f_{n-1}(x)$.Differentiating both sides with respect to $x$:$\frac{1}{f_n(x)} \cdot f_n'(x) = f_{n-1}'(x)$$\Rightarrow f_n'(x) = f_n(x) \cdot f_{n-1}'(x) \quad \dots (i)$For $n=1$, $f_1'(x) = e^x = f_1(x)$.For $n=2$, $f_2'(x) = f_2(x) \cdot f_1'(x) = f_2(x) \cdot f_1(x)$.For $n=3$, $f_3'(x) = f_3(x) \cdot f_2'(x) = f_3(x) \cdot f_2(x) \cdot f_1(x)$.By induction, for any $n \geq 1$, $\frac{d}{dx} f_n(x) = f_n(x) \cdot f_{n-1}(x) \cdot \ldots \cdot f_1(x)$.

Let $f_1(x) = e^x, f_2(x) = e^{f_1(x)}, \ldots, f_{n+1}(x) = e^{f_n(x)}$ for all $n \geq 1$. Then for any fixed $n$, $\frac{d}{dx} f_n(x)$ is:

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