## Condition of the problem

Find the derivative $y_{x}^{\prime}$.

$$
\left\{\begin{array}{l}
x=\ln \sqrt{\frac{1-\sin t}{1+\sin t}} \\
y=\frac{1}{2} \operatorname{tg}^{2} t+\ln \cos t
\end{array}\right.
$$
## Solution

$$
\begin{aligned}
& x_{t}^{\prime}=\left(\ln \sqrt{\frac{1-\sin t}{1+\sin t}}\right)^{\prime}=\sqrt{\frac{1+\sin t}{1-\sin t}} \cdot\left(\sqrt{\frac{1-\sin t}{1+\sin t}}\right)^{\prime}= \\
& =\sqrt{\frac{1+\sin t}{1-\sin t}} \cdot \frac{1}{2 \sqrt{\frac{1-\sin t}{1+\sin t}}} \cdot\left(\frac{1-\sin t}{1+\sin t}\right)^{\prime}= \\
& =\sqrt{\frac{1+\sin t}{1-\sin t}} \cdot \frac{1}{2} \cdot \sqrt{\frac{1+\sin t}{1-\sin t}} \cdot\left(\frac{(1-\sin t)^{\prime} \cdot(1+\sin t)-(1-\sin t) \cdot(1+\sin t)^{\prime}}{(1+\sin t)^{2}}\right)= \\
& =\frac{1}{2} \cdot \frac{1+\sin t}{1-\sin t} \cdot\left(\frac{-\cos t \cdot(1+\sin t)-(1-\sin t) \cdot \cos t}{(1+\sin t)^{2}}\right)= \\
& =\frac{1}{2(1-\sin t)} \cdot \frac{-\cos t-\cos t \cdot \sin t-\cos t+\sin t \cdot \cos t}{1+\sin t}= \\
& =\frac{-2 \cos t}{2\left(1-\sin ^{2} t\right)}=\frac{-\cos t}{\cos ^{2} t}=-\frac{1}{\cos t} \\
& y_{t}^{\prime}=\left(\frac{1}{2} \operatorname{tg}^{2} t+\ln \cos t\right)^{\prime}=\frac{1}{2} \cdot 2 \cdot(\operatorname{tg} t)^{\prime}+\frac{1}{\cos t} \cdot(\cos t)^{\prime}= \\
& =\frac{1}{\cos ^{2} t}+\frac{1}{\cos ^{\prime} t} \cdot(-\sin t)=\frac{1-\sin t \cdot \cos t}{\cos ^{2} t}
\end{aligned}
$$

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{\frac{1-\sin t \cdot \cos t}{\cos ^{2} t}}{-\frac{1}{\cos t}}=\frac{1-\sin t \cdot \cos t}{\cos ^{2} t} \cdot(-\cos t)= \\
& =\frac{\sin t \cdot \cos t-1}{\cos t}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 16-25