Answer
$y’=\frac{1}{\ln 10}\left(\frac{x^2+1}{x(x^2-1)}\right)$
Work Step by Step
$y=\frac{\ln (\frac{x^2-3}{x})}{\ln 10}$
$=\frac{\ln (x^2-1)-\ln x}{\ln 10}$
$y’=\frac{1}{\ln 10}\left(\frac{2x}{x^2-1}-\frac{1}{x}\right)$
$y’=\frac{1}{\ln 10}\left(\frac{x^2+1}{x(x^2-1)}\right)$
