91Ó°ÊÓ

The chain rule is a fundamental tool in calculus for finding derivatives of composite functions. In our exercise, we are given the function \( y = (1 + 2x)^{10} \), which is a composition of the outer function \( u^{10} \) and the inner function \( u = 1 + 2x \). The chain rule provides a systematic approach to differentiate such functions.

To apply the chain rule, we differentiate the outer function, treating the inner function as a single variable, and then multiply it by the derivative of the inner function. In mathematical terms, if \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is given by:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

For our problem, the derivative of the outer function, \( (1+2x)^{10} \), is \( 10(1+2x)^9 \) and the derivative of the inner function \( 1+2x \) is \( 2 \). So, the chain rule yields:
\[ \frac{dy}{dx} = 10(1 + 2x)^9 \cdot 2 = 20(1 + 2x)^9 \]
This simplified process makes finding derivatives of more complex functions manageable.

In calculus, a derivative represents the rate of change of a function with respect to a variable. In simpler terms, it describes how a function changes as its input changes. For a curve's tangent line, the derivative at a specific point gives the slope of the tangent line at that point.

For the function \( y = (1 + 2x)^{10} \), calculating the derivative helps us determine how the curve behaves around any point. Using the chain rule, the derivative we found is \( \frac{dy}{dx} = 20(1 + 2x)^9 \). This expression tells us the rate at which \( y \) changes as \( x \) changes.

• The derivative is key in applications across physics, engineering, and economics, wherever change is analyzed.
• It provides insights into a function's behavior, such as where it increases or decreases.

This is why understanding the derivative is crucial for tackling tangent line problems.

The equation of a line provides a way to represent a linear relationship graphically. In two-dimensional space, a line can be described using the equation \( y - y_1 = m(x - x_1) \), known as the point-slope form.

In our exercise, we need the tangent line at the point \((0, 1)\) on the curve. The slope \( m \) is already calculated as 20. Using the point-slope form equation, we substitute point \((0, 1)\) and slope \( m = 20 \) to get:

• \( y - 1 = 20(x - 0) \)

By simplifying, we find the line equation:
\[ y = 20x + 1 \]
This equation now represents the tangent line at the point \((0, 1)\). It's a simple and effective way to understand how a straight line can closely approximate a curve at a specific point.

Slope is the measure of steepness of a line, defined as the ratio of vertical change to horizontal change between two points. For a tangent line at a particular point on a curve, the slope indicates how the curve behaves precisely at that point.

In the exercise problem, after using the chain rule, the derivative \( \frac{dy}{dx} = 20(1+2x)^9 \) gives us a formula for the slope at any point \( x \). To find the slope of the tangent line at the point \((0, 1)\), we substitute \( x = 0 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=0} = 20(1 + 2 \times 0)^9 = 20(1)^9 = 20 \]

• This gives a slope of 20 for the tangent line at the point \((0, 1)\).

Slope calculations are fundamental in understanding geometry and physics, as they determine how angles and directions are established on a graph.