## Condition of the problem

Find the differential $d y$

$y=\ln \left(\tan \frac{x}{2}\right)-\frac{x}{\sin x}$
## Solution

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

## Problem Kuznetsov Differentiation 4-14