## Task Condition

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

$$
\left\{\begin{array}{l}
x=\ln (\operatorname{ctg} t) \\
y=\frac{1}{\cos ^{2} t}
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=(\ln (\operatorname{ctg} t))^{\prime}=\frac{1}{\operatorname{ctg} t} \cdot \frac{-1}{\sin ^{2} t}=-\frac{1}{\cos t \cdot \sin t}$

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

We obtain:

$$
y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{2 \sin t}{\cos ^{3} t}\right) /\left(-\frac{1}{\cos t \cdot \sin t}\right)=-\frac{2 \sin t \cdot \cos t \cdot \sin t}{\cos ^{3} t}=-2 \operatorname{tg}^{2} t
$$

## Problem Kuznetsov Differentiation 16-8