Problem 43 - Applied Calculus

Question

Decide which function is an antiderivative of the other. $$f(x)=1-\frac{1}{x^{2}} ; g(x)=\frac{1}{x}+x$$

Step-by-Step Solution

Answer

The function \( g(x) = \frac{1}{x} + x \) is an antiderivative of \( f(x) = 1 - \frac{1}{x^2} \).

Step 1: Understand the Problem

We are given two functions, \( f(x) = 1 - \frac{1}{x^2} \) and \( g(x) = \frac{1}{x} + x \). We need to find out if one is an antiderivative of the other.

Step 2: Find the Derivative of \( g(x) \)

Calculate the derivative of the function \( g(x) = \frac{1}{x} + x \). To differentiate \( g(x) \), apply the power rule to \( x \) (derivative is 1) and the derivative of \( \frac{1}{x} \) which is \( -\frac{1}{x^2} \).Thus, \( g'(x) = -\frac{1}{x^2} + 1 \).

Step 3: Compare \( f(x) \) and \( g'(x) \)

Notice that \( f(x) = 1 - \frac{1}{x^2} \) and \( g'(x) = -\frac{1}{x^2} + 1 \). These two expressions are equivalent, though the terms are ordered differently.Therefore, \( f(x) \) is the same as \( g'(x) \).

Step 4: Conclusion

Since \( f(x) = g'(x) \), \( g(x) \) is an antiderivative of \( f(x) \).

Key Concepts

Function Differentiation

Function Differentiation is a core concept in calculus. It involves finding the derivative of a function, which is essentially the rate at which the function is changing at any given point.
To differentiate a function, you look at how its output changes as you make tiny changes in the input. This helps in understanding the behavior of the function.
In the given exercise, the goal was to differentiate the function \( g(x) = \frac{1}{x} + x \). By finding its derivative, we can understand how quickly it changes with respect to \( x \). The process of differentiation allows us to transform a function into its derivative, providing insight into the slope or steepness of the curve represented by the function.

• The derivative of \( \frac{1}{x} \) was found to be \(-\frac{1}{x^2}\).
• The derivative of \( x \) was more straightforward and is \( 1 \).

Understanding differentiation is crucial, as it lays the foundation for finding antiderivatives and working with more complex functions.

Power Rule

The Power Rule is a handy shortcut in calculus used to differentiate functions of the form \( x^n \). According to the Power Rule, the derivative of \( x^n \) is \( n \cdot x^{n-1} \).
This rule is particularly useful when dealing with polynomial functions or expressions that can be rewritten in terms of powers of \( x \).
In the exercise given, we applied the Power Rule when differentiating the individual terms in \( g(x) \).

• For \( x \), which is \( x^1 \), its derivative is \( 1 \cdot x^{1-1} = 1 \).
• For \( \frac{1}{x} \), which can be expressed as \( x^{-1} \), the derivative becomes \(-1 \cdot x^{-1-1} = -x^{-2} = -\frac{1}{x^2} \).

Using the Power Rule simplifies the calculation process, making it easier and less time-consuming to find derivatives of power functions.

Comparing Functions

After obtaining the derivative, comparing functions is a crucial step to determine relationships, such as whether one function is an antiderivative of another.
In the exercise, the functions to be compared were \( f(x) = 1 - \frac{1}{x^2} \) and \( g'(x) = -\frac{1}{x^2} + 1 \). By carefully examining their expressions, we can see that they are mathematically equivalent.

• Notice how both functions contain \(-\frac{1}{x^2}\) and a constant term \(1\).
• The positioning of the terms doesn't affect their equivalence, as addition is commutative.

Recognizing this equivalence confirms that \( f(x) \) is indeed the derivative of \( g(x) \), indicating that \( g(x) \) is an antiderivative of \( f(x) \). This comparison is vital in mathematics, allowing us to simplify problems and identify deeper connections between functions.