If y = log^n x,where log^n denotes the n-th i | 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} = \frac{d}

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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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