This problem appears to ask for something related to calculus, possibly the computation of a derivative, tangent line, or some integral around the point where $y = 1$. The expressions given involve $y = \sqrt{x}$, and the notation "about $y = 1$" suggests we're analyzing behavior near that point. Let's go step-by-step.

Given:

• $y = \sqrt{x}$
• $x = 0$ and $y = 1$, which implies we're examining around $y = 1$.

To analyze the problem, I'll assume it's about finding the slope of the tangent line to the curve $y = \sqrt{x}$ at $x = 1$, since at that point, $y = 1$.

1. Find the derivative of $y = \sqrt{x}$:

Since $y = \sqrt{x}$ can be rewritten as $y = x^{1/2}$, we differentiate it with respect to $x$:

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

2. Evaluate the derivative at $x = 1$:

When $x = 1$, $y = \sqrt{1} = 1$, and the derivative becomes:

$\frac{dy}{dx}\bigg|_{x=1} = \frac{1}{2\sqrt{1}} = \frac{1}{2}.$

So, the slope of the tangent line at $x = 1$ is $\frac{1}{2}$.

3. Equation of the tangent line:

The equation of the tangent line to the curve at the point $(1, 1)$ can be written as:

$y - y_1 = m(x - x_1),$

where $m = \frac{1}{2}$ (the slope) and $(x_1, y_1) = (1, 1)$. Substituting the values:

$y - 1 = \frac{1}{2}(x - 1),$

which simplifies to:

$y = \frac{1}{2}x + \frac{1}{2}.$

4. Behavior near $y = 1$:

Around $y = 1$, the tangent line approximates the curve $y = \sqrt{x}$, and the line equation $y = \frac{1}{2}x + \frac{1}{2}$ provides a linear approximation.

Conclusion:

The derivative of $y = \sqrt{x}$ at $x = 1$ is $\frac{1}{2}$, and the equation of the tangent line near $y = 1$ is $y = \frac{1}{2}x + \frac{1}{2}$. This approximation holds near the point $y = 1$.

Would you like further details on this solution? Let me know!

Here are 5 related questions to expand on this topic:

1. How do we find the tangent line to a curve at any arbitrary point?
2. What is the geometric interpretation of the derivative at a point?
3. How would the analysis change if we looked at the behavior near a different point on the curve?
4. What happens to the derivative of $y = \sqrt{x}$ as $x \to 0$?
5. How can we use Taylor series to approximate $y = \sqrt{x}$ around $x = 1$?

Tip: In calculus, understanding the behavior of a function near specific points often requires examining both the function's derivative and the tangent line, which provide useful approximations for small changes around the point of interest.