## Condition of the problem

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

$$
\left\{\begin{array}{l}
x=e^{\sec ^{2} t} \\
y=\operatorname{tg} t \cdot \ln \cos t+\operatorname{tg} t-t
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=\left(e^{\sec ^{2} t}\right)^{\prime}=\left(e^{\frac{1}{\cos ^{2} t}}\right)^{\prime}=e^{\frac{1}{\cos ^{2} t}} \cdot\left(\frac{1}{\cos ^{2} t}\right)^{\prime}=e^{\frac{1}{\cos ^{2} t}} \cdot \frac{-2}{\cos ^{3} t} \cdot(-\sin t)=$ $=e^{\frac{1}{\cos ^{2} t}} \cdot \frac{2 \sin t}{\cos ^{3} t}$

$y_{t}^{\prime}=(\tan t \cdot \ln \cos t+\tan t-t)^{\prime}=\frac{1}{\cos ^{2} t} \cdot \ln \cos t+\tan t \cdot \frac{1}{\cos t} \cdot(-\sin t)+\frac{1}{\cos ^{2} t}-1=$ $=\frac{\ln \cos t}{\cos ^{2} t}-\frac{\sin ^{2} t}{\cos ^{2} t}+\frac{1}{\cos ^{2} t}-\frac{\cos ^{2} t}{\cos ^{2} t}=\frac{\ln \cos t}{\cos ^{2} t}+\frac{-\sin ^{2} t+1-\cos ^{2} t}{\cos ^{2} t}=$
$=\frac{\ln \cos t}{\cos ^{2} t}+\frac{-1+1}{\cos ^{2} t}=\frac{\ln \cos t}{\cos ^{2} t}$

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{\ln \cos t}{\cos ^{2} t}\right) /\left(e^{\frac{1}{\cos ^{2} t}} \cdot \frac{2 \sin t}{\cos ^{3} t}\right)=\frac{\ln \cos t}{\cos ^{2} t} \cdot e^{-\frac{1}{\cos ^{2} t}} \cdot \frac{\cos ^{3} t}{2 \sin t}= \\
& =\frac{\cos t \cdot \ln \cos t}{2 \sin t} \cdot e^{-\sec ^{2} t}=\frac{1}{2} \cdot \cot t \cdot \ln \cos t \cdot e^{-\sec ^{2} t}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 16-29