## Problem Statement

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

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

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

We obtain:

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

Then:

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

## Problem Kuznetsov Differentiation 20-12