Question

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

Answer 1

$y=\tan ^{-1} \frac{\log \left(e / x^{2}\right)}{\log \left(e x^{2}\right)}+\tan ^{-1} \frac{3+2 \log x}{1-6 \log x}$
$=\tan ^{-1} \frac{\left(\log e-\log x^{2}\right)}{\left(\log e+\log x^{2}\right)}+\tan ^{-1} \frac{3+2 \log x}{1-6 \log x}$
$=\tan ^{-1}\left(\frac{1-2 \log x}{1+2 \log x}\right)+\tan ^{-1} 3+\tan ^{-1} \log x^{2}$
$=\tan ^{-1}(1)-\tan ^{-1} \log \,x ^{2}+\tan ^{-1} 3+\tan ^{-1} \log x^{2}$
$=\tan ^{-1}(1)+\tan ^{-1}(3)$
$\frac{d^{2} y}{d x}=0$