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The derivative of a function measures how the function changes as its input changes. This is essential for understanding the behavior of curves and their tangents. When we look at the graph of a function like \( y = \frac{1}{x} \), the derivative helps us determine the slope of the tangent line at any given point on the curve.

• The "slope" is simply a measure of how steep the curve is at any point.
• For \( y = \frac{1}{x} \), the derivative is \( y' = -\frac{1}{x^2} \).

This derivative tells us that, as \( x \) increases, the slope becomes less negative, and conversely, as \( x \) decreases, the slope becomes more negative.Understanding derivatives is crucial because they give us the equation of the tangent. Knowing the equation of the tangent is key to forming triangles with other lines, such as the coordinate axes.

Coordinate geometry involves plotting points, lines, and curves on a coordinate plane, allowing us to analyze and solve geometric problems algebraically.In our exercise, once we have the derivative, it helps us form a tangent to the curve \( y = \frac{1}{x} \). The equation of the tangent line is derived using the point-slope formula, which is fundamental to coordinate geometry:

• Point-slope form is given by \( y - y_1 = m(x - x_1) \).
• For any point \((x_0, y_0)\) on our curve, \( m \) is \(-\frac{1}{x_0^2}\).
• The tangent line equation simplifies to \( y = -\frac{1}{x_0^2}x + \frac{2}{x_0} \).

Using this form, we can find where this line intersects the coordinate axes. These intercepts are crucial as they form the base and height of the triangle in our area calculation.

The area of a triangle is an invaluable concept in geometry, calculated as \( \frac{1}{2} \times \text{base} \times \text{height} \). By understanding the positions of the triangle's vertices, we can easily compute its area.For our specific problem concerning the tangent line:

• The x-intercept is \((2x_0, 0)\).
• The y-intercept is \((0, \frac{2}{x_0})\).
• The origin \((0, 0)\) is also a vertex.

The triangle thus formed has its base along the x-axis from \(0\) to \(2x_0\) and its height equal to \( \frac{2}{x_0} \). The vertices form a right-angle triangle, simplifying our calculation:\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2x_0 \times \frac{2}{x_0} = 2 \]This straightforward calculation shows that, no matter where the tangent touches the curve, the area of the resulting triangle remains constant at 2 square units.