## Problem Statement

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

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

$x_{t}^{\prime}=(t+\sin t)^{\prime}=1+\cos t$

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

We obtain:

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{-\sin t}{1+\cos t}=-\frac{\sin t}{1+\cos t}$

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

$=-\frac{\cos t+\cos ^{2} t+\sin ^{2} t}{(1+\cos t)^{2}}=-\frac{\cos t+1}{(1+\cos t)^{2}}=-\frac{1}{1+\cos t}$

Then:

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

## Problem Kuznetsov Differentiation 20-19