## Problem Statement

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

$\left\{\begin{array}{l}x=e^{t} \\ y=\arcsin t\end{array}\right.$
## Solution

$x_{t}^{\prime}=\left(e^{t}\right)^{\prime}=e^{t}$

$y_{t}^{\prime}=(\arcsin t)^{\prime}=\frac{1}{\sqrt{1-t^{2}}}$

We obtain:

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

$$
\begin{aligned}
& =-\frac{1}{e^{t} \cdot\left(1-t^{2}\right)} \cdot\left(\sqrt{1-t^{2}}-\frac{t}{\sqrt{1-t^{2}}}\right)= \\
& =-\frac{1}{e^{t} \cdot\left(1-t^{2}\right)} \cdot \frac{1-t^{2}-t}{\sqrt{1-t^{2}}}=\frac{t^{2}+t-1}{e^{t} \cdot \sqrt{\left(1-t^{2}\right)^{3}}}
\end{aligned}
$$

Then:

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

## Problem Kuznetsov Differentiation 20-23