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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.

Solve a calculus problem involving the derivative of y = √x at x = 1, the slope of the tangent line, and linear approximation near y = 1. Learn step-by-step how to calculate the slope, tangent line, and behavior of the curve around this point.

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