If y=tan ^{-1}left(frac{log left(frac{e}{x^2} | vedclass

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(A) Given $y=	an ^{-1}\left(\frac{\log \left(\frac{e}{x^2}ight)}{\log \left(ex^2ight)}ight)+	an ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}ight

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(A) Given $y=	an ^{-1}\left(\frac{\log \left(\frac{e}{x^2}ight)}{\log \left(ex^2ight)}ight)+	an ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}ight)$.Using properties of logarithms,$\log(\frac{e}{x^2}) = \log e - \log x^2 = 1 - 2\log x$ and $\log(ex^2) = \log e + \log x^2 = 1 + 2\log x$.So,$y = 	an^{-1}\left(\frac{1-2\log x}{1+2\log x}ight) + 	an^{-1}\left(\frac{4+2\log x}{1-8\log x}ight)$.Using $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we have $	an^{-1}(1) - 	an^{-1}(2\log x) = 	an^{-1}\left(\frac{1-2\log x}{1+2\log x}ight)$.Also,$	an^{-1}(4) + 	an^{-1}(2\log x) = 	an^{-1}\left(\frac{4+2\log x}{1-4(2\log x)}ight) = 	an^{-1}\left(\frac{4+2\log x}{1-8\log x}ight)$.Substituting these back,$y = 	an^{-1}(1) - 	an^{-1}(2\log x) + 	an^{-1}(4) + 	an^{-1}(2\log x)$.$y = 	an^{-1}(1) + 	an^{-1}(4)$.Since $y$ is a constant,$\frac{dy}{dx} = 0$.

If $y=\tan ^{-1}\left(\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e x^2\right)}\right)+\tan ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\right)$,then $\frac{d y}{d x}$ is

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