点 $P$ 沿方程为 $y = x \sin^{-1} x$（$-1 < x < 1$）的曲线 $C$ 移动。设曲线 $C$ 在点 $P$ 处的斜率为 $m$。 如果在 $x = \frac{1}{2}$ 时，点 $P$ 的 $x$ 坐标以每秒 9 个单位的速度增加，求在此瞬间 $m$ 变化率的精确值。

1. Differentiate $y = x\sin^{-1}x$ to find an expression for the gradient $m$.

Using the product rule:
$m = \frac{dy}{dx} = (1)\sin^{-1} x + x \left( \frac{1}{\sqrt{1-x^2}} \right)$
$m = \sin^{-1} x + \frac{x}{\sqrt{1-x^2}}$!

2. Find $\frac{dm}{dx}$ by differentiating $m$ with respect to $x$.

Differentiate $m$ with respect to $x$ using the quotient rule for the second term:
$\frac{dm}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{(1)\sqrt{1-x^2} - x\left(\frac{1}{2\sqrt{1-x^2}}\right)(-2x)}{1-x^2}$
$\frac{dm}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{\sqrt{1-x^2} + \frac{x^2}{\sqrt{1-x^2}}}{1-x^2}$
$\frac{dm}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{1-x^2+x^2}{(1-x^2)^{3/2}}$
$\frac{dm}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{1}{(1-x^2)^{3/2}} = \frac{1-x^2}{(1-x^2)^{3/2}} + \frac{1}{(1-x^2)^{3/2}} = \frac{2-x^2}{(1-x^2)^{3/2}}$

3. Use the connected rates of change formula $\frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt}$ at $x = \frac{1}{2}$.

Substitute $x = \frac{1}{2}$ into the expression for $\frac{dm}{dx}$:
$\frac{dm}{dx} = \frac{2 - \left(\frac{1}{2}\right)^2}{\left(1 - \left(\frac{1}{2}\right)^2\right)^{3/2}} = \frac{\frac{7}{4}}{\left(\frac{3}{4}\right)^{3/2}} = \frac{\frac{7}{4}}{\frac{3\sqrt{3}}{8}} = \frac{14}{3\sqrt{3}}$
Given that $\frac{dx}{dt} = 9$, apply the chain rule:
$\frac{dm}{dt} = \frac{dm}{dx} \times \frac{dx}{dt}$
$\frac{dm}{dt} = \frac{14}{3\sqrt{3}} \times 9 = \frac{126}{3\sqrt{3}} = \frac{42}{\sqrt{3}} = 14\sqrt{3}$