The point $P$ travels along the curve $C$ with equation $y = x \sin^{-1} x$, $-1 < x < 1$. Let the gradient of the curve $C$ at the point $P$ be $m$. If the $x$-coordinate of $P$ is increasing at the rate of 9 units per second when $x = \frac{1}{2}$, find the exact value of the rate at which $m$ is changing at this instant.

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:
$\frac{dm}{dt} = \frac{dm}{dx} \times \frac{dx}{dt}$
$\frac{dm}{dt} = \frac{14}{3\sqrt{3}} \times 9 = \frac{126}{3\sqrt{3}} = \frac{42}{\sqrt{3}} = 14\sqrt{3}$