Chapter 2 - Differentiation - 2.3 Exercises - Page 125: 48

$y'=\dfrac{(\sec(x))(x\tan(x)-1)}{x^2}$.

Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=\sec(x); u'(x)=\sec(x)\tan(x)$ $v(x)=x; v'(x)=1$ $y'=\dfrac{(\sec(x)\tan(x))(x)-(1)(\sec(x))}{x^2}$ $=\dfrac{(\sec(x))(x\tan(x)-1)}{x^2}$