## Problem Statement

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

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

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

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

We obtain:

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{t \cdot \sin t}{t \cdot \cos t}=\frac{\sin t}{\cos t}=\operatorname{tg} t$

$\left(y_{x}^{\prime}\right)_{t}^{\prime}=(\operatorname{tg} t)^{\prime}=\frac{1}{\cos ^{2} t}$

Then:

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

## Problem Kuznetsov Differentiation 20-22