If f(x)=cot ^{-1}left(frac{x^x-x^{-x}}{2}righ | vedclass

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(D) Given,$f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}ight)$.Let $y = \cot ^{-1}\left(\frac{x^{2x}-1}{2x^x}ight)$.Put $x^x = 	an 	heta$,then $y =

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-$1$

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(D) Given,$f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}ight)$.Let $y = \cot ^{-1}\left(\frac{x^{2x}-1}{2x^x}ight)$.Put $x^x = 	an 	heta$,then $y = \cot ^{-1}\left(\frac{	an^2 	heta - 1}{2 	an 	heta}ight)$.Since $\cot 2	heta = \frac{1-	an^2 	heta}{2 	an 	heta}$,we have $y = \cot ^{-1}(-\cot 2	heta)$.Using the property $\cot ^{-1}(-z) = \pi - \cot ^{-1}(z)$,we get $y = \pi - \cot ^{-1}(\cot 2	heta) = \pi - 2	heta$.Substituting back $	heta = 	an ^{-1}(x^x)$,we have $y = \pi - 2 	an ^{-1}(x^x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = -2 \cdot \frac{1}{1+(x^x)^2} \cdot \frac{d}{dx}(x^x)$.Since $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$,we have $\frac{dy}{dx} = -\frac{2x^x(1 + \ln x)}{1 + x^{2x}}$.At $x=1$,$\frac{dy}{dx} = -\frac{2(1)^1(1 + \ln 1)}{1 + (1)^2} = -\frac{2(1)(1+0)}{2} = -1$.

If $f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$,then $f^{\prime}(1)=$

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