## Problem Statement

Find the derivative.

$$
y=\frac{(1+x) \operatorname{arctg} \sqrt{x}}{x^{2}}+\frac{1}{3 x \sqrt{x}}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{(1+x) \operatorname{arctg} \sqrt{x}}{x^{2}}+\frac{1}{3 x \sqrt{x}}\right)^{\prime}=\left(\frac{(1+x) \operatorname{arctg} \sqrt{x}}{x^{2}}+\frac{1}{3} \cdot x^{-\frac{3}{2}}\right)^{\prime}= \\
& =\frac{((1+x) \operatorname{arctg} \sqrt{x})^{\prime} \cdot x^{2}-(1+x) \operatorname{arctg} \sqrt{x} \cdot\left(x^{2}\right)^{\prime}}{x^{4}}+\frac{1}{3} \cdot\left(-\frac{3}{2}\right) \cdot x^{-\frac{5}{2}}= \\
& =\frac{\left(\operatorname{arctg} \sqrt{x}+(1+x)(\operatorname{arctg} \sqrt{x})^{\prime}\right) \cdot x^{2}-(1+x) \operatorname{arctg} \sqrt{x} \cdot 2 x}{x^{4}}-\frac{1}{2} \cdot x^{-\frac{5}{2}}= \\
& =\frac{x^{2} \cdot \operatorname{arctg} \sqrt{x}+x^{2} \cdot(1+x) \cdot \frac{1}{1+(\sqrt{x})^{2}} \cdot(\sqrt{x})^{\prime}-2 x \cdot \operatorname{arctg} \sqrt{x}-2 x^{2} \cdot \operatorname{arctg} \sqrt{x}}{x^{4}}-\frac{1}{2} \cdot x^{-\frac{5}{2}}= \\
& =\frac{x^{2} \cdot(1+x) \cdot \frac{1}{1+x} \cdot \frac{1}{2 \sqrt{x}}-2 x \cdot \operatorname{arctg} \sqrt{x}-x^{2} \cdot \operatorname{arctg} \sqrt{x}}{x^{4}}-\frac{1}{2} \cdot \frac{1}{\sqrt{x^{5}}}= \\
& =\frac{x \sqrt{x} \cdot \frac{1}{2}-x \cdot(2+x) \cdot \operatorname{arctg} \sqrt{x}}{x^{4}}-\frac{1}{2} \cdot \frac{1}{x^{2} \cdot \sqrt{x}}=\frac{\sqrt{x}-2(2+x) \cdot \operatorname{arctg} \sqrt{x}}{2 x^{3}}-\frac{\sqrt{x}}{2 x^{3}}= \\
& =\frac{\sqrt{x}-2(2+x) \cdot \operatorname{arctg} \sqrt{x}-\sqrt{x}}{2 x^{3}}=-\frac{2(2+x) \cdot \operatorname{arctg} \sqrt{x}}{2 x^{3}}=-\frac{(2+x) \cdot \operatorname{arctg} \sqrt{x}}{x^{3}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 10-9