Answer
$ \displaystyle \frac{1}{2\sqrt{x}}(\sqrt{x}+\frac{1}{x^{2}})+(\sqrt{x}+1)(\frac{1}{2\sqrt{x}}-\frac{2}{x^{3}})$
Work Step by Step
$(f\cdot g)^{\prime}=f^{\prime}g+fg^{\prime}$
...written in exponential form,
$f(x)=x^{1/2}+1,\displaystyle \qquad f^{\prime}(x)=\frac{1}{2}x^{-1/2}$
$g(x)=x^{1/2}+x^{-2},\displaystyle \qquad g^{\prime}(x)=\frac{1}{2}x^{-1/2}-2x^{-3}$
$\displaystyle \frac{dy}{dx}=(f\cdot g)^{\prime}(x)=$
$=(\displaystyle \frac{1}{2}x^{-1/2})(x^{1/2}+x^{-2}) +(x^{1/2}+1)(\frac{1}{2}x^{-1/2}-2x^{-3})$
...back to radical form...
$=\displaystyle \frac{1}{2\sqrt{x}}(\sqrt{x}+\frac{1}{x^{2}})+(\sqrt{x}+1)(\frac{1}{2\sqrt{x}}-\frac{2}{x^{3}})$
... we do not need to expand the answer..