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

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

Aicharya · Accepted Answer

$\begin{aligned} & y=	an ^{-1}\left(\frac{\log \left(\frac{\mathrm{e}}{x^2}ight)}{\log \left(\mathrm{ex}^2ight)}ight)+	an ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}ight) \ & =	an ^{-1}\left(\frac{\log \mathrm{e}-\log x^2}{\log \mathrm{e}+\log x^2}ight)+	an ^{-1}(4)+	an ^{-1}(2 \log x) \ & =	an ^{-1}\left(\frac{1-2 \log x}{1+2 \log x}ight)+	an ^{-1}(4)+	an ^{-1}(2 \log x) \ & \quad=	an ^{-1}(1)-	an ^{-1}(2 \log x)+	an ^{-1}(4)+	an ^{-1}(2 \log x) \ & 	herefore \quad y=	an ^{-1}(1)+	an ^{-1}(4) \ & 	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=0\end{aligned}$

If $y=\tan ^{-1}\left(\frac{\log \left(\frac{\mathrm{e}}{x^2}\right)}{\log \left(e x^2\right)}\right)+\tan…

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