一个质点沿着曲线 $y = \frac{2}{\sqrt{3x+1}}$ 运动，其 $y$ 坐标以每秒 3 个单位的恒定速率减少。求当 $x$ 坐标以每秒 0.125 个单位的速率增加的瞬间，该质点的 $y$ 坐标。 证明当 $x > \frac{1}{e}$ 时，$y = x \ln x^2$ 是一个递增函数。

1. Use the chain rule to find the value of $\frac{dy}{dx}$ at the given instant.

By the chain rule, $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$.
Given $\frac{dy}{dt} = -3$ and $\frac{dx}{dt} = 0.125$,
$-3 = \frac{dy}{dx} \times 0.125$
$\frac{dy}{dx} = \frac{-3}{0.125} = -24$

2. Differentiate the equation of the curve with respect to $x$.

$y = 2(3x+1)^{-\frac{1}{2}}$
$\frac{dy}{dx} = 2 \left(-\frac{1}{2}\right)(3x+1)^{-\frac{3}{2}} \times 3$
$\frac{dy}{dx} = -3(3x+1)^{-\frac{3}{2}}$

3. Equate the two expressions for $\frac{dy}{dx}$ to solve for $x$.

$-3(3x+1)^{-\frac{3}{2}} = -24$
$(3x+1)^{-\frac{3}{2}} = 8$
$(3x+1)^{\frac{3}{2}} = \frac{1}{8}$
$3x+1 = \left(\frac{1}{8}\right)^{\frac{2}{3}}$
$3x+1 = \frac{1}{4}$
$3x = -\frac{3}{4}$
$x = -\frac{1}{4}$

4. Substitute the value of $x$ into the original equation to find the $y$-coordinate.

When $x = -\frac{1}{4}$,
$y = \frac{2}{\sqrt{3\left(-\frac{1}{4}\right)+1}}$
$y = \frac{2}{\sqrt{\frac{1}{4}}}$
$y = \frac{2}{\frac{1}{2}} = 4$

1. Differentiate the function with respect to $x$.

Rewrite $y = x \ln x^2$ as $y = 2x \ln x$ (since $x > 0$).
Using the product rule:
$\frac{dy}{dx} = 2 \ln x + 2x \left(\frac{1}{x}\right)$
$\frac{dy}{dx} = 2 \ln x + 2$

2. Show that the derivative is positive for the given domain.

For $y$ to be an increasing function, we need $\frac{dy}{dx} > 0$.
Given $x > \frac{1}{e}$, we have $x > e^{-1}$.
Taking the natural logarithm on both sides (since $\ln x$ is an increasing function):
$\ln x > \ln(e^{-1})$
$\ln x > -1$
Multiplying by 2 and adding 2:
$2 \ln x > -2$
$2 \ln x + 2 > 0$
Since $\frac{dy}{dx} > 0$ for all $x > \frac{1}{e}$, $y = x \ln x^2$ is an increasing function in this domain.