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If $y=\tan ^{-1}\left(\frac{\log \left(\frac{\mathrm{e}}{x^2}\right)}{\log \left(e x^2\right)}\right)+\tan ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\right)$, then $\frac{d y}{d x}$ is
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$\begin{aligned} & y=\tan ^{-1}\left(\frac{\log \left(\frac{\mathrm{e}}{x^2}\right)}{\log \left(\mathrm{ex}^2\right)}\right)+\tan ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\right) \\ & =\tan ^{-1}\left(\frac{\log \mathrm{e}-\log x^2}{\log \mathrm{e}+\log x^2}\right)+\tan ^{-1}(4)+\tan ^{-1}(2 \log x) \\ & =\tan ^{-1}\left(\frac{1-2 \log x}{1+2 \log x}\right)+\tan ^{-1}(4)+\tan ^{-1}(2 \log x) \\ & \quad=\tan ^{-1}(1)-\tan ^{-1}(2 \log x)+\tan ^{-1}(4)+\tan ^{-1}(2 \log x) \\ & \therefore \quad y=\tan ^{-1}(1)+\tan ^{-1}(4) \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=0\end{aligned}$
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