Answer
$\frac{1}{\sqrt (x^{2}-1)}+arcsecx$
Work Step by Step
$y=xarcsecx$
Through the product rule of derivatives and the derivative of $arcsecu=\frac{\frac{du}{dx}}{|u|\sqrt (u^{2}-1)}$:
$\frac{dy}{dx}=x(\frac{1}{|x|\sqrt (x^{2}-1)})+arcsecx$
$\frac{dy}{dx}=\frac{1}{\sqrt (x^{2}-1)}+arcsecx$
