## Condition of the problem

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

$$
\left\{\begin{array}{l}
x=\operatorname{ctg}\left(2 e^{t}\right) \\
y=\ln \left(\operatorname{tg} e^{t}\right)
\end{array}\right.
$$
## Solution

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

We obtain:

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{2 e^{t}}{\sin \left(2 e^{t}\right)}\right) /\left(\frac{-2 e^{t}}{\sin ^{2}\left(2 e^{t}\right)}\right)=-\sin \left(2 e^{t}\right)$

## Problem Kuznetsov Differentiation 16-7