## Task Condition

Find the differential $d y$.

$$
y=\ln |\cos \sqrt{x}|+\sqrt{x} \tan \sqrt{x}
$$
## Solution

$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=(\ln |\cos \sqrt{x}|+\sqrt{x} \tan \sqrt{x})^{\prime} d x= \\
& =\left(\frac{1}{\cos \sqrt{x}} \cdot(-\sin \sqrt{x}) \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x}} \tan \sqrt{x}+\sqrt{x} \cdot \frac{1}{\cos ^{2} \sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}\right) d x=
\end{aligned}
$$

$$
=\left(-\tan \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x}} \tan \sqrt{x}+\frac{1}{\cos ^{2} \sqrt{x}} \cdot \frac{1}{2}\right) d x=\frac{d x}{2 \cos ^{2} \sqrt{x}}
$$

## Problem Kuznetsov Differentiation 4-23