Example 6. $y=\operatorname{arctg} \sqrt{x}$.
Solution. We apply rule 5), formulas 14) and 16). Let $u=\sqrt{x}$. Then $y=\operatorname{arctg} u, y_{u}^{\prime}=\frac{1}{1+u^{2}}, u^{\prime}=\frac{1}{2 \sqrt{x}}$. Finally: $y_{u}^{\prime}=\frac{1}{2 \sqrt{x}(1+x)}$.