## Task Condition

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

$$
\left\{\begin{array}{l}
x=\operatorname{arctg} t \\
y=\frac{t^{2}}{2}
\end{array}\right.
$$
## Solution

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

$y_{t}^{\prime}=\left(\frac{t^{2}}{2}\right)^{\prime}=t$

We obtain:

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

Then:

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

## Problem Kuznetsov Differentiation 20-26