## Task Condition

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

$$
\left\{\begin{array}{l}
x=\cos 2 t \\
y=2 \sec ^{2} t
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=(\cos 2 t)^{\prime}=-2 \sin 2 t$

$y_{t}^{\prime}=\left(2 \sec ^{2} t\right)^{\prime}=\left(\frac{2}{\cos ^{2} t}\right)^{\prime}=-\frac{4}{\cos ^{3} t} \cdot(-\sin t)=\frac{4 \sin t}{\cos ^{3} t}$

We obtain:

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

$=-\frac{4 \sin t}{4 \cos ^{4} t \cdot \sin t}=-\frac{1}{\cos ^{4} t}$

$\left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(-\frac{1}{\cos ^{4} t}\right)^{\prime}=-\frac{-4}{\cos ^{5} t} \cdot(-\sin t)=-\frac{4 \sin t}{\cos ^{5} t}$

$$
\begin{aligned}
& y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{4 \sin t}{\cos ^{5} t}\right) /(-2 \sin 2 t)=\frac{4 \sin t}{2 \cos ^{5} t \cdot \sin 2 t}= \\
& =\frac{4 \sin t}{4 \cos ^{6} t \cdot \sin t}=\frac{1}{\cos ^{6} t}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 20-1