Derivative of $${\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)^2}$$   is :

Solution :
$$\eqalign{ & {\text{We have, }}\frac{d}{{dx}}\left\{ {{{\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)}^2}} \right\} \cr & = \frac{d}{{dx}}\left\{ {x + \frac{1}{x} + 2} \right\} \cr & = \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {{x^{ - 1}}} \right) + \frac{d}{{dx}}\left( 2 \right) \cr & = 1 + \left( { - 1} \right){x^{ - 2}} + 0 \cr & = 1 - \frac{1}{{{x^2}}} \cr} $$