6.38. Find the second derivative of the function given parametrically $x(t)=\ln t, y(t)=t^{3}+2 t-3+7$.

---

Note: The original text had a minor redundancy in the function $y(t)$, where $-3+7$ can be simplified to $+4$. However, I have kept the expression as it was in the original text.
Solution.

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{i}^{\prime}}=\frac{\left(t^{3}+2 t-3\right)_{t}^{\prime}}{(\ln t)_{t}^{\prime}}=\frac{3 t^{2}+2}{\frac{1}{t}}=3 t^{3}+2 t, \\
& y_{x}^{\prime \prime}=\frac{\left(y_{t}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\frac{\left(3 t^{3}+2 t\right)_{t}^{\prime}}{(\ln t)_{t}^{\prime}}=\frac{9 t^{2}+2}{\frac{1}{t}}=9 t^{3}+2 t .
\end{aligned}
$$