91Ó°ÊÓ

Differentiation is a fundamental concept in calculus, primarily used to understand how a function changes at any given point. When you differentiate a function, you are essentially finding its derivative. The derivative provides the rate of change or the slope of the tangent line to the curve at any particular point.
For instance, with the function given as \( y = 1 + 2 \sin x \), the task is to find \( \frac{dy}{dx} \), which is the derivative of \( y \) with respect to \( x \). In simpler terms, we want to determine how \( y \) changes for a tiny change in \( x \).
The process used here involves applying basic differentiation rules:

• The derivative of a constant, such as 1, is 0.
• The derivative of \( \sin x \) with respect to \( x \) is \( \cos x \).
• Due to the constant multiplication rule, the derivative of \( 2 \sin x \) becomes \( 2 \cos x \).

Putting these together, we find \( \frac{dy}{dx} = 2 \cos x \). This expression gives us the slope of the tangent at any point on the curve described by \( y = 1 + 2 \sin x \).

The sine function, \( \sin x \), is a periodic function that oscillates between -1 and 1. It is one of the primary trigonometric functions, and it plays a critical role in describing wave-like phenomena. Understanding the sine function is essential for applications in physics, engineering, and other scientific fields.
When we have an equation like \( y = 1 + 2 \sin x \), the sine function is being scaled and shifted:

• The "2" in front of the sine stretches its amplitude, making the wave oscillate between -2 and 2, but considering the constant "1" added to it, it oscillates between -1 and 3.
• The addition of "1" shifts the entire sine wave upwards by 1 unit, altering its mean position.

In this exercise, the sine function helps form a curve along which a tangent line is drawn. By substituting \( x = \frac{\pi}{6} \) into \( y = 1 + 2 \sin x \), we determine the exact point where the tangent meets the curve, resulting in the coordinates \( \left(\frac{\pi}{6}, 2\right) \).

The point-slope form of a linear equation is a valuable formula for writing the equation of a line when you know a point on the line and its slope. It is expressed as \( y - y_1 = m(x - x_1) \), where:

• \( (x_1, y_1) \) represents a point on the line.
• \( m \) is the slope of the line.

In our exercise, after finding the derivative at \( x = \frac{\pi}{6} \), we evaluated it to get the slope \( \sqrt{3} \). We also found the point of tangency to be \( \left(\frac{\pi}{6}, 2\right) \).
By plugging these values into the point-slope form, we arrive at the equation of the tangent line:
\( y - 2 = \sqrt{3} \left(x - \frac{\pi}{6}\right) \).
This equation indicates how the slope \( \sqrt{3} \) interacts with the curve at precisely the touchpoint. The point-slope form is particularly useful for visualizing and graphing the tangent line next to the curve.