91Ó°ÊÓ

Calculus, a branch of mathematics, helps us analyze how things change. It is particularly focused on two main ideas, differentiation and integration. Differentiation is the process of finding the rate at which a function is changing at any point, often represented through the concept of a derivative.
Integration, on the other hand, lets us understand the accumulation of quantities. In this exercise, we're primarily dealing with differentiation. By using derivatives, we can find tangent lines - these lines touch a curve at a point without crossing over. Understanding this property is crucial for solving problems that involve rates of change and slopes at specific points on a curve.

• Calculus helps describe the nature of curves and their rates of change.
• In the problem, using calculus allows us to derive the slope of the tangent line to the curve at the given point.

The derivative of a function gives us a lot of important information. It allows us to find the slope of the tangent line at any given point on a curve. When we talk about a derivative, we are essentially asking: How rapidly and in what manner does this function change?
To find the derivative of the function in this exercise, we employed a very common rule called the power rule. This rule makes it easy to take derivatives of functions in the form of powers of x.
For example:

• The function \( y = \sqrt{x} \) can be rewritten as \( y = x^{1/2} \).
• Using the power rule, the derivative \( y' \) becomes \( \frac{1}{2}x^{-1/2} \).

This gives the slope of the tangent line at any point \( x \) on the curve.
For this exercise, evaluating the derivative at the point \( x = 1 \), provided us with the slope \( m = \frac{1}{2} \). This slope is key to forming the equation of the tangent line.

The point-slope form is a useful method for finding the equation of a line when a point on the line and the slope is known. It is particularly handy in calculus problems that involve derivatives and tangent lines.
The formula for the point-slope form is:

• \( y - y_1 = m(x - x_1) \)

where \((x_1, y_1)\) is the point on the line, and \(m\) is the slope.
For our problem:

• The point is \((1,1)\) and the slope \(m\) is \(\frac{1}{2}\).

Plugging these values into the point-slope form gives us:

• \( y - 1 = \frac{1}{2}(x - 1) \)

This equation can be further simplified to express \( y \) in terms of \( x \), resulting in the final tangent line equation \( y = \frac{1}{2}x + \frac{1}{2} \). This form helps us understand exactly how a particular point on a curve locally behaves.