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The chain rule is essential when finding the derivative of a composite function. In other words, it helps us differentiate functions that are nested within each other. Suppose we are given the function \( y = \ln(x^2) \), which is a combination of the natural logarithm \( \ln \) and the square function \( x^2 \). Here's how the chain rule works in this scenario.

• First, identify the outer function \( \ln(u) \) with \( u = x^2 \).
• The derivative of the outer function, \( \ln(u) \), with respect to \( u \) is \( \frac{1}{u} \).
• The inner function, \( u = x^2 \), has a derivative of \( 2x \).

Using the chain rule, the derivative of \( \ln(x^2) \) becomes \( \frac{1}{u} \cdot 2x = \frac{2}{x} \). This shows how the two functions combine and helps us find the slope of the tangent line at any point on the curve. It’s important to master the chain rule as it is widely applicable in calculus, particularly for differentiating complex functions.

The area of a triangle is a straightforward calculation once you know the base and height. The formula is \( A = \frac{1}{2} \times \text{base} \times \text{height} \). In the given problem, the challenge was to find the area of a triangle made from the tangent, a horizontal line through point \( P \), and the y-axis.
To find the area, follow these steps:

• Determine the base as the vertical difference between the intersections of the triangle with the \( y \)-axis. In this case, the base was calculated as \( 2 \) using the points \( (0, 2\ln w - 2) \) and \( (0, 2\ln w) \).
• Find the height, which is the horizontal distance from the y-axis to point \( P(w, 2\ln w) \). Here, the height is \( w \).

Substitute these values into the area formula to get \( A = \frac{1}{2} \times 2 \times w \). Simplifying this gives \( A(w) = w \). This method of finding area is crucial when working with coordinate geometry and calculus.

The point-slope form of a line's equation is a powerful tool in calculus and algebra. When you have a point on a line and the slope of the line, you can easily write the equation of the line. The standard format is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is a known point on the line.
Here's how it's applied in the problem:

• The known point on the tangent line is \( P(w, \ln(w^2)) \).
• The slope \( m \) comes from the derivative \( \frac{2}{w} \).
• Substituting into the point-slope formula gives \( y - \ln(w^2) = \frac{2}{w}(x - w) \).

This form quickly leads to the equation \( y = \frac{2}{w}x - 2 + 2\ln w \), which represents the tangent line passing through \( P \). Utilizing the point-slope form is essential, especially when dealing with tangent lines and other applications requiring linear equations in calculus.