## Problem Statement

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

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

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

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

We obtain:

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

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

Kuznetsov Differentiation Problem 20-18