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

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

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(D) Given, $f(x)=	an ^{-1}\left[\frac{\log \left(\frac{e}{x^{2}}ight)}{\log \left(e x^{2}ight)}ight]+	an ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}ight]$Using the properties of logarithms, $\log(e/x^2) = \log e - 2 \log x = 1 - 2 \log x$ and $\log(ex^2) = \log e + 2 \log x = 1 + 2 \log x$.So, $f(x) = 	an^{-1}\left[\frac{1 - 2 \log x}{1 + 2 \log x}ight] + 	an^{-1}\left[\frac{3 + 2 \log x}{1 - 3(2 \log x)}ight]$.Let $2 \log x = u$. Then $f(x) = 	an^{-1}\left[\frac{1 - u}{1 + u}ight] + 	an^{-1}\left[\frac{3 + u}{1 - 3u}ight]$.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)$:$f(x) = (	an^{-1} 1 - 	an^{-1} u) + (	an^{-1} 3 + 	an^{-1} u)$.$f(x) = \frac{\pi}{4} + 	an^{-1} 3$.Since $f(x)$ is a constant, its derivative $f'(x) = 0$ and the second derivative $f''(x) = 0$.

If $f(x)=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^{2}}\right)}{\log \left(e x^{2}\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$ then the value of $f^{\prime \prime}(x)$ is equal to

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