If y=sec ^{-1}left(frac{x+x^{-1}}{x-x^{-1}}ri | vedclass

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Question

(B) Given $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$.Simplify the expression inside the $\sec^{-1}$ function:$\frac{x+x^{-1}}{x-x^{-1}} = \f

Vedclass · Accepted Answer

$\frac{-2}{1+x^2}$

Vedclass · Answer

(B) Given $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}ight)$.Simplify the expression inside the $\sec^{-1}$ function:$\frac{x+x^{-1}}{x-x^{-1}} = \frac{x+\frac{1}{x}}{x-\frac{1}{x}} = \frac{x^2+1}{x^2-1} = -\frac{1+x^2}{1-x^2}$.Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.The expression becomes $-\frac{1+	an^2 	heta}{1-	an^2 	heta} = -\frac{1}{\cos 2	heta} = -\sec 2	heta$.So,$y = \sec^{-1}(-\sec 2	heta)$.Using the identity $\sec^{-1}(-z) = \pi - \sec^{-1}(z)$,we get $y = \pi - \sec^{-1}(\sec 2	heta) = \pi - 2	heta$.Substituting back $	heta = 	an^{-1} x$,we have $y = \pi - 2	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\pi) - 2\frac{d}{dx}(	an^{-1} x) = 0 - 2\left(\frac{1}{1+x^2}ight) = -\frac{2}{1+x^2}$.

If $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$,then $\frac{d y}{d x}=$

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