## Problem Statement

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

$$
\left\{\begin{array}{l}
x=\cos t \\
y=\ln (\sin t)
\end{array}\right.
$$
## Solution

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

$y_{t}^{\prime}=(\ln (\sin t))^{\prime}=\frac{1}{\sin t} \cdot \cos t=\frac{\cos t}{\sin t}$

We obtain:

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

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

$=\frac{\sin ^{2} t+2 \cos ^{2} t}{\sin ^{3} t}=\frac{1+\cos ^{2} t}{\sin ^{3} t}$

Then:

$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{1+\cos ^{2} t}{\sin ^{3} t}\right) /(-\sin t)=-\frac{1+\cos ^{2} t}{\sin ^{4} t}$

## Problem Kuznetsov Differentiation 20-21