## Task Condition

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

$$
\left\{\begin{array}{l}
x=\sqrt{t-3} \\
y=\ln (t-2)
\end{array}\right.
$$
## Solution

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

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

We obtain:

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

Then:

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

$=-\frac{2 t \sqrt{t-1}}{\sqrt{t-1} \cdot(t-2)^{2}}=-\frac{2 t}{(t-2)^{2}}$

## Problem Kuznetsov Differentiation 20-17