## Problem Statement

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

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

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

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

We obtain:

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

Then:
$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2(\cos t-\sin t)}{-\sin t+\cos t}=2$

## Problem Kuznetsov Differentiation 20-30