1. We are asked to find the derivative of the function $$y = \arcsin\left(\frac{x}{6}\right)$$ and then evaluate it at $$x=4$$. 2. Recall the derivative formula for $$y = \arcsin(u)$$ where $$u$$ is a function of $$x$$: $$\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$$ 3. Here, $$u = \frac{x}{6}$$, so we first find $$\frac{du}{dx}$$: $$\frac{du}{dx} = \frac{d}{dx}\left(\frac{x}{6}\right) = \frac{1}{6}$$ 4. Substitute $$u$$ and $$\frac{du}{dx}$$ into the derivative formula: $$\frac{dy}{dx} = \frac{1}{\sqrt{1-\left(\frac{x}{6}\right)^2}} \cdot \frac{1}{6} = \frac{1}{6\sqrt{1-\frac{x^2}{36}}}$$ 5. Simplify the expression under the square root: $$1 - \frac{x^2}{36} = \frac{36 - x^2}{36}$$ 6. Substitute back: $$\frac{dy}{dx} = \frac{1}{6 \sqrt{\frac{36 - x^2}{36}}} = \frac{1}{6} \cdot \frac{1}{\frac{\sqrt{36 - x^2}}{6}} = \frac{1}{6} \cdot \frac{6}{\sqrt{36 - x^2}}$$ 7. Cancel the 6's: $$\frac{dy}{dx} = \frac{\cancel{1}}{\cancel{6}} \cdot \frac{\cancel{6}}{\sqrt{36 - x^2}} = \frac{1}{\sqrt{36 - x^2}}$$ 8. Now evaluate at $$x=4$$: $$\frac{dy}{dx}\bigg|_{x=4} = \frac{1}{\sqrt{36 - 4^2}} = \frac{1}{\sqrt{36 - 16}} = \frac{1}{\sqrt{20}}$$ 9. Simplify $$\sqrt{20}$$: $$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$$ 10. Final answer: $$\frac{dy}{dx}\bigg|_{x=4} = \frac{1}{2\sqrt{5}}$$