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

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(D) Given $y=	an ^{-1}\left[\frac{\log \left(\frac{e}{x^2}ight)}{\log \left(ex^2ight)}ight]+	an ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}ight

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(D) Given $y=	an ^{-1}\left[\frac{\log \left(\frac{e}{x^2}ight)}{\log \left(ex^2ight)}ight]+	an ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}ight]$.Using properties of logarithms,$\log(e/x^2) = \log e - \log x^2 = 1 - 2\log x$ and $\log(ex^2) = \log e + \log x^2 = 1 + 2\log x$.Let $u = 2\log x$. Then the first term is $	an^{-1}\left(\frac{1-u}{1+u}ight) = 	an^{-1}(1) - 	an^{-1}(u) = \frac{\pi}{4} - 	an^{-1}(2\log x)$.For the second term,$	an^{-1}\left[\frac{3+2 \log x}{1-6 \log x}ight] = 	an^{-1}(3) + 	an^{-1}(2\log x)$.Adding these,$y = \frac{\pi}{4} - 	an^{-1}(2\log x) + 	an^{-1}(3) + 	an^{-1}(2\log x) = \frac{\pi}{4} + 	an^{-1}(3)$.Since $y$ is a constant,$\frac{dy}{dx} = 0$ and $\frac{d^2 y}{dx^2} = 0$.

If $y=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(ex^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$,then $\frac{d^2 y}{dx^2}=$

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