## Task Condition

Find the derivative.

$y=\frac{1}{2 \sqrt{x}}+\frac{1+x}{2 x} \cdot \operatorname{arctg} \sqrt{x}$
## Solution

$y^{\prime}=\left(\frac{1}{2 \sqrt{x}}+\frac{1+x}{2 x} \cdot \operatorname{arctg} \sqrt{x}\right)^{\prime}=$
$=-\frac{1}{2} \cdot \frac{1}{2 \sqrt{x^{3}}}+\frac{1 \cdot x-(1+x) \cdot 1}{2 x^{2}} \cdot \operatorname{arctg} \sqrt{x}+\frac{1+x}{2 x} \cdot \frac{1}{1+(\sqrt{x})^{2}} \cdot \frac{1}{2 \sqrt{x}}=$

$=-\frac{1}{4 x \sqrt{x}}+\frac{-1}{2 x^{2}} \cdot \operatorname{arctg} \sqrt{x}+\frac{1+x}{2 x} \cdot \frac{1}{1+x} \cdot \frac{1}{2 \sqrt{x}}=$

$=-\frac{1}{4 x \sqrt{x}}-\frac{1}{2 x^{2}} \cdot \operatorname{arctg} \sqrt{x}+\frac{1}{4 x \sqrt{x}}=-\frac{1}{2 x^{2}} \cdot \operatorname{arctg} \sqrt{x}$

## Problem Kuznetsov Differentiation 10-11