frac{d}{d x}left(cos ^{-1}left(frac{x-frac{1} | vedclass

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(D) Let $y = \cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)$.Simplifying the expression inside the inverse cosine function:$y = \cos ^{-1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $y = \cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}ight)$.Simplifying the expression inside the inverse cosine function:$y = \cos ^{-1}\left(\frac{x^2-1}{x^2+1}ight) = \cos ^{-1}\left(-\left(\frac{1-x^2}{1+x^2}ight)ight)$.Using the property $\cos ^{-1}(-z) = \pi - \cos ^{-1}(z)$,we get:$y = \pi - \cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight)$.Substitute $x = 	an 	heta$,which implies $	heta = 	an ^{-1} x$:$y = \pi - \cos ^{-1}\left(\frac{1-	an ^2 	heta}{1+	an ^2 	heta}ight)$.Since $\cos 2	heta = \frac{1-	an ^2 	heta}{1+	an ^2 	heta}$,we have:$y = \pi - \cos ^{-1}(\cos 2	heta) = \pi - 2	heta = \pi - 2	an ^{-1} x$.Differentiating with respect to $x$:$\frac{d y}{d x} = \frac{d}{d x}(\pi - 2	an ^{-1} x) = 0 - 2\left(\frac{1}{1+x^2}ight) = \frac{-2}{1+x^2}$.

$\frac{d}{d x}\left(\cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\right)=$

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