## Task Condition

Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.

$$
\left\{\begin{array}{l}
x=\tan t \\
y=\frac{1}{\sin 2 t}
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=(\tan t)^{\prime}=\frac{1}{\cos ^{2} t}$

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

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{2 \cos 2 t}{\sin ^{2} 2 t}\right) /\left(\frac{1}{\cos ^{2} t}\right)=-\frac{2 \cos 2 t \cdot \cos ^{2} t}{4 \sin ^{2} t \cdot \cos ^{2} t}=-\frac{\cos 2 t}{2 \sin ^{2} t}= \\
& =-\frac{1-2 \sin ^{2} t}{2 \sin ^{2} t}=-\left(\frac{1}{2 \sin ^{2} t}-1\right)=1-\frac{1}{2 \sin ^{2} t} \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(1-\frac{1}{2 \sin ^{2} t}\right)^{\prime}=-\frac{-2}{2 \sin ^{3} t} \cdot \cos t=\frac{\cos t}{\sin ^{3} t}
\end{aligned}
$$

Then:

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

$$
=-\frac{4 \sin ^{2} t \cdot \cos ^{3} t}{2 \sin ^{3} t \cdot \cos 2 t}=-\frac{2 \cdot \cos ^{3} t}{\sin t \cdot \cos 2 t}
$$

## Problem Kuznetsov Differentiation 20-9