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

Answer: 0.00

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