91Ó°ÊÓ

In calculus, derivatives are fundamental in understanding how a function changes at any given point. They represent the rate of change or slope of the function. For the function given in the problem, \[ y = \ln x^2 \]can be rewritten using logarithm properties as \[ y = 2\ln x \].This simplifies solving for derivatives and recognizing the function’s behavior.

The derivative of this function, often denoted as \( \frac{dy}{dx} \),represents the tangent’s slope at any point \( x \).Finding the derivative involves taking the derivative of \( 2\ln x \),which is simply \( \frac{2}{x} \).This derivative helps us find the specific slope needed to create equations that illustrate changes at points on the curve.

Remember:

• Derivatives are concerned with the slope of a function.
• The derivative of a natural logarithm often uses chain rules or property simplifications.
• This slope aids in creating equations of tangents, such as lines that intersect the function only at a single point.

Once you have the derivative, creating the tangent line equation is the next step. This equation describes the line that just touches the curve at a single point, without crossing it. In the given problem, the tangent line at point \((w, 2\ln w)\)can be found using the point-slope form of a line:
\[ y - y_1 = m(x - x_1) \].

Here, \( m \)is the slope calculated from the derivative \( \frac{2}{w} \),and \((x_1, y_1)\)is the point of tangency \((w, 2\ln w)\).By substituting these into the equation, you get:\[ y - 2\ln w = \frac{2}{w}(x - w) \].

This simplifies to \( y = \frac{2x}{w} - 2 \),which is the equation of the tangent line. Understanding this step reinforces how derivatives and tangent lines relate directly through slope, intercepts, and their point of contact with a curve.

The final part involves calculating the area of a triangle formed by a tangent line, a horizontal line, and the y-axis. Considering the triangle defined in the problem:
1. The base is fully along the y-axis between points \((0, 2\ln w)\)and \((0, -2)\).2. The total height is \(2\ln w + 2\),available by calculating the vertical distance between these two points.

• Your horizontal line supports the base parallel to the x-axis.
• The tangent line ensures completion of the triangle by intersecting vertically.

The formula for the area \( A(w) = \frac{1}{2} \times \text{base} \times \text{height} \)applies. With the base \( w \)spanning this formula, the area calculates to:\[ A(w) = w(\ln w + 1) \].Understanding this step emphasizes how intersections define dimensional constraints, determining the triangle's position and size, particularly within calculus challenges and applications.