## Task Condition

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

$$
\left\{\begin{array}{l}
x=\ln t \\
y=\operatorname{arctg} t
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=(\ln t)^{\prime}=\frac{1}{t}$

$y_{t}^{\prime}=(\operatorname{arctg} t)^{\prime}=\frac{1}{1+t^{2}}$

We obtain:

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

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

Kuznetsov Differentiation Problem 20-31