If $y = \log^{n}{x,}$where $\log^{n}{means\ }$log log log…(repeated $n$ times), then $x\log{x\log^{2}x}\log^{3}x\ldots\log^{n - 1}{x\log^{n}{x\frac{dy}{dx}}}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
$\because y = \log^{n}x$
On differentiating w.r.t. $x$, we get
$x\log{x\log^{2}{x\log^{3}{x\ldots\log^{n - 1}{x\log^{n}{x\frac{dy}{dx}}}}}}$
$= \frac{x\log{x\log^{2}{x\log^{3}{x\ldots\log^{n - 1}{x\log^{n}{x.1}}}}}}{x\log{x\log^{2}{x\log^{3}{x\ldots\log^{n - 1}x}}}}$
$= \log^{n}x$

Q. If $y = \log^{n}{x,}$where $\log^{n}{means\ }$log log log…(repeated $n$ times), then $x\log{x\log^{2}x}\log^{3}x\ldots\log^{n - 1}{x\log^{n}{x\frac{dy}{dx}}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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