## Condition of the problem

Find the differential $d y$.

$$
y=\cos x \cdot \ln (\operatorname{tg} x)-\ln \left(\operatorname{tg} \frac{x}{2}\right)
$$
## Solution

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

## Problem Kuznetsov Differentiation 4-27