If y = log^n x, where log^n denotes the n-th | vedclass

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(C) Let $y_n = \log^n x$. Then $y_n = \log(y_{n-1})$, where $y_1 = \log x$, $y_2 = \log(\log x)$, and so on.By the chain rule, $\frac{dy_n}{dx} = \fra

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(C) Let $y_n = \log^n x$. Then $y_n = \log(y_{n-1})$, where $y_1 = \log x$, $y_2 = \log(\log x)$, and so on.By the chain rule, $\frac{dy_n}{dx} = \frac{d}{dx}(\log y_{n-1}) = \frac{1}{y_{n-1}} \cdot \frac{dy_{n-1}}{dx}$.Expanding this, we get $\frac{dy_n}{dx} = \frac{1}{y_{n-1}} \cdot \frac{1}{y_{n-2}} \cdot \dots \cdot \frac{1}{y_1} \cdot \frac{d}{dx}(\log x)$.Since $\frac{d}{dx}(\log x) = \frac{1}{x}$, we have $\frac{dy_n}{dx} = \frac{1}{y_{n-1} y_{n-2} \dots y_1 x}$.Substituting $y_k = \log^k x$, we get $\frac{dy}{dx} = \frac{1}{x \log x \log^2 x \dots \log^{n-1} x}$.Therefore, $x \log x \log^2 x \dots \log^{n-1} x \frac{dy}{dx} = 1$.

If $y = \log^n x$, where $\log^n$ denotes the $n$-th iterated logarithm $\log_e(\log_e(\dots \log_e x \dots))$ ($n$ times), then $x \log x \log^2 x \log^3 x \dots \log^{n-1} x \log^n x \frac{dy}{dx}$ is equal to

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