To find the slope of the tangent line to the curve $y = \sqrt{x}$ at the point where $x = 8$, we can use the derivative that we previously calculated.

Steps:

1. Derivative of $y = \sqrt{x}$: As we found earlier, the derivative of $y = \sqrt{x}$ is: $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$

2. Evaluate the derivative at $x = 8$: Substitute $x = 8$ into the derivative to find the slope at this point: $\frac{dy}{dx}\bigg|_{x=8} = \frac{1}{2\sqrt{8}} = \frac{1}{2 \times \sqrt{8}} = \frac{1}{2 \times 2\sqrt{2}} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8}$

   Simplifying further: $\frac{\sqrt{2}}{8} \approx 0.177$

So, the slope of the tangent line to the curve at $x = 8$ is approximately $0.177$.

Would you like to explore this further or have any additional questions?

Here are five related questions to expand your understanding:

1. What is the exact value of the slope at $x = 2$ for the same curve?
2. How would you compute the slope of the tangent for a different function, say $y = x^2$?
3. How does the slope change as $x$ increases along the curve $y = \sqrt{x}$?
4. Can you find the slope of the tangent line at $x = 1$ and interpret it geometrically?
5. What happens to the slope as $x$ approaches 0 for the curve $y = \sqrt{x}$?

Tip: Always express your final slope as both an exact value and, when possible, an approximation to make it easier to interpret in practical scenarios.