To solve the problem of finding the slope of the tangent line to the curve $y = x^{1/2}$ at the point $(1, 1)$, we will first estimate the slope graphically and then verify the result analytically.

Step 1: Estimation using the graph

We know that the graph of $y = x^{1/2}$ is the same as $y = \sqrt{x}$, which is a familiar shape — a curve that increases more slowly as $x$ grows.

At the point $(1, 1)$, the graph is smooth, and the slope of the tangent line represents the instantaneous rate of change of the function at that point. From visual inspection of such a graph (without an actual visual provided here), we expect the tangent line at $(1, 1)$ to have a slope around 0.5 to 1, as the curve is less steep near $x = 1$.

However, to be precise, we will verify this using calculus.

Step 2: Verifying analytically

The slope of the tangent line at any point on a curve is given by the derivative of the function at that point.

The function we are given is:

$y = x^{1/2}$

Now, let's differentiate this function using the power rule. The derivative of $x^n$ is $n \cdot x^{n-1}$.

$\frac{dy}{dx} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$

Now, substitute $x = 1$ to find the slope at the point $(1, 1)$:

$\frac{dy}{dx} \Bigg|_{x=1} = \frac{1}{2\sqrt{1}} = \frac{1}{2} = 0.5$

Step 3: Conclusion

Thus, the slope of the tangent line to the curve $y = x^{1/2}$ at the point $(1, 1)$ is 0.5, both by estimation and by analytical verification.

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Would you like any further clarification or details?

Here are five related questions to deepen understanding:

1. What is the derivative of $y = x^n$ for any $n$?
2. How does the slope of the tangent line change as you move along the curve $y = x^{1/2}$?
3. What is the geometric interpretation of the derivative at a point?
4. How do you find the equation of the tangent line once you have the slope?
5. What is the slope of the tangent line to $y = x^{1/3}$ at $(1,1)$?

Tip: Always check the units of the derivative to ensure you're interpreting the slope correctly.