## Task Condition

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

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

$y_{x}^{\prime}=\frac{y^{\prime}(t)}{x^{\prime}(t)}$

$y^{\prime}(t)=\left(1 / \sin ^{2} t\right)^{\prime}=-\frac{2 \sin t \cdot \cos t}{\sin ^{4} t}=-\frac{\sin 2 t}{\sin ^{4} t}$

$x^{\prime}(t)=(\ln \operatorname{tg} t)^{\prime}=\frac{1}{\operatorname{tg} t} \cdot(\operatorname{tg} t)^{\prime}=\frac{\cos t}{\sin t} \cdot \frac{1}{\cos ^{2} t}=\frac{1}{\cos t \cdot \sin t}$

$y_{x}^{\prime}=-\frac{\sin 2 t}{\sin ^{4} t} \cdot \cos t \cdot \sin t=-\frac{2 \sin t \cdot \cos t \cdot \cos t}{\sin ^{3} t}=-2 \operatorname{ctg}^{2} t$.

## Problem Kuznetsov Differentiation 16-27