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

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(C) Let $u = \log_e x$. Then the expression becomes: $y = 	an^{-1}\left[\frac{\log_e e - \log_e x^2}{\log_e e + \log_e x^2}ight] + 	an^{-1}\left[\

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(C) Let $u = \log_e x$. Then the expression becomes: $y = 	an^{-1}\left[\frac{\log_e e - \log_e x^2}{\log_e e + \log_e x^2}ight] + 	an^{-1}\left[\frac{3+2u}{1-6u}ight]$ $y = 	an^{-1}\left[\frac{1-2u}{1+2u}ight] + 	an^{-1}\left[\frac{3+2u}{1-3(2u)}ight]$ Using the formula $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$ and $	an^{-1} A + 	an^{-1} B = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$: $y = (	an^{-1} 1 - 	an^{-1}(2u)) + (	an^{-1} 3 + 	an^{-1}(2u))$ $y = 	an^{-1} 1 + 	an^{-1} 3$ Since $y$ is a constant,the derivative $\frac{dy}{dx} = 0$. Therefore,the second derivative $\frac{d^2 y}{d x^2} = 0$.

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

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