91Ó°ÊÓ

In the world of calculus, the **derivative** of a function is crucial in understanding how the function behaves. Simply put, the derivative at any point on a function gives us the rate of change or the slope of the tangent line at that point.
For the function \( f(x) = \ln|x| \), the derivative is found using calculus rules, specifically focusing on the [natural logarithm function](https://en.wikipedia.org/wiki/Natural_logarithm).
The derivative of \( \ln|x| \) is \( f'(x) = \frac{1}{x} \). This formula is derived from the basic rules of differentiation where \( \frac{d}{dx} \ln|x| \) leads to this expression due to the chain rule in handling absolute values.
To find the slope of the tangent line at a specific point \( x_0 = -2 \), we substitute \( -2 \) into our derivative to get \( f'(-2) = -\frac{1}{2} \). This tells us how steeply the line rises or falls at that point.

The **natural logarithm function**, denoted as \( \ln(x) \), is a core mathematical function that provides the power to which a constant base, called Euler's number \( e \approx 2.718 \, \text{(a constant value)} \), must be raised to produce the number \( x \). It's an important function in calculus and is often encountered when dealing with exponential growth or decay.
In our problem, the function is given as \( f(x) = \ln|x| \), where the absolute value handles both positive and negative values of \( x \). The natural logarithm of a positive number \( x \) is just \( \ln(x) \), while the natural logarithm involving negative numbers requires the absolute value, simplifying our function to \( \ln|x| \).
At \( x_0 = -2 \), we evaluate the function and obtain \( f(-2) = \ln(2) \). This step involves finding how the position on the graph relates to the \( x \) and \( y \) coordinates, resulting in the point \((-2, \ln(2))\).

The **point-slope form** of a line equation is a straightforward method to write the equation of a line when you're given the slope and a point on the line. It's commonly formulated as:

• \( y - y_1 = m(x - x_1) \)

where \( m \) represents the slope and \( (x_1, y_1) \) is the point through which the line passes.
In our tangent line problem, we already determined that the slope \( m \) is \(-\frac{1}{2}\) and the point of tangency is \( (-2, \ln(2)) \).
By inserting these values into the point-slope equation, we get:
\[ y - \ln(2) = -\frac{1}{2}(x + 2) \]
This equation perfectly represents the tangent line at \( x = -2 \), describing how the line touches the graph of \( f(x) = \ln|x| \) at that exact point.