If y = b cos log left( frac{x}{n} right)^n,th | vedclass

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(C) Given $y = b \cos \log \left( \frac{x}{n} \right)^n$. Using the chain rule,we differentiate with respect to $x$: $\frac{dy}{dx} = -b \sin \left( \

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$\frac{- nb}{x} \sin \log \left( \frac{x}{n} \right)^n$

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(C) Given $y = b \cos \log \left( \frac{x}{n} \right)^n$. Using the chain rule,we differentiate with respect to $x$: $\frac{dy}{dx} = -b \sin \left( \log \left( \frac{x}{n} \right)^n \right) \cdot \frac{d}{dx} \left( \log \left( \frac{x}{n} \right)^n \right)$. Using the property of logarithms,$\log \left( \frac{x}{n} \right)^n = n \log \left( \frac{x}{n} \right)$. So,$\frac{d}{dx} \left( n \log \left( \frac{x}{n} \right) \right) = n \cdot \frac{1}{x/n} \cdot \frac{1}{n} = \frac{n}{x}$. Therefore,$\frac{dy}{dx} = -b \sin \left( \log \left( \frac{x}{n} \right)^n \right) \cdot \frac{n}{x} = -\frac{nb}{x} \sin \log \left( \frac{x}{n} \right)^n$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

If $y = b \cos \log \left( \frac{x}{n} \right)^n$,then $\frac{dy}{dx} = $

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