91Ó°ÊÓ

The power rule is a fundamental tool in calculus, making differentiation much simpler. It tells us how to find the derivative of a power of a variable. For any function of the form \(x^n\), the derivative is found using the formula:

• \(\frac{d}{dx}(x^n) = nx^{n-1}\)

This means that we multiply the power by the coefficient in front of the \(x\), and then reduce the power by 1. This approach applies to all terms in a polynomial.

In the problem, for \(F(x) = 5x^3 + 2x^2 + 3x + 1\), the derivatives are solved using the power rule:

• The derivative of \(5x^3\) is \(15x^2\) because \(3\times5 = 15\) and \(x^{3-1} = x^2\).
• The derivative of \(2x^2\) is \(4x\) because \(2\times2 = 4\) and \(x^{2-1} = x\).
• The derivative of \(3x\) is \(3\) since the power is reduced to \(0\), leaving the coefficient.
• A constant term like \(1\) differentiates to \(0\) because it has no variable to change.

These calculations show that applying the power rule leads directly to finding the derivative, which in this case means verifying the antiderivative relationship between \(F(x)\) and \(f(x)\).

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change of a function concerning its variable. Differentiation transforms a function, making it possible to examine things like slope, velocity, etc.

In the context of the exercise, differentiating \(F(x) = 5x^3 + 2x^2 + 3x + 1\) using the power rule, helps us verify the relationship between \(F(x)\) and \(f(x)\). By differentiating, we find:

• \(F'(x) = 15x^2 + 4x + 3\)

This result shows how changes in the curve of \(F(x)\) relate directly to \(f(x)\). By successfully differentiating \(F(x)\) to derive \(f(x)\), we confirm \(F(x)\)'s role as an antiderivative. This exemplifies the connection between finding derivatives of polynomial expressions and identifying their antiderivatives.

The Fundamental Theorem of Calculus is an essential concept linking differentiation and integration. This theorem connects the antiderivative and the definite integral, showing that integrating a function \(f(x)\) can "undo" differentiation.

The theorem comprises two parts:

• The first part states that if \(f(x)\) is continuous on an interval \([a, b]\), then there exists an antiderivative \(F(x)\) such that \(F'(x) = f(x)\).
• The second part provides a way to evaluate definite integrals using these antiderivatives: \(\int_a^b f(x) dx = F(b) - F(a)\).

In this problem, the differentiation of \(F(x)\) ending with \(f(x)\) illustrates the first part of the theorem. It shows \(F(x)\) as an antiderivative of \(f(x)\). By showing that \(F'(x)\) matches \(f(x)\), the exercise validates the initial definition of \(F(x)\) as an antiderivative, bringing to life this crucial aspect of calculus.