Problem 71 - Calculus with Concepts in Calculus

Question

Let \(F\) and \(G\) be antiderivatives of \(f\) and \(g\), respectively, on an interval \(I\). a. Prove that \(F+G\) is an antiderivative of \(f+g\) on \(I\). b. Prove that \(c F\) is an antiderivative of \(c f\) on \(I\) for any constant \(c\). c. Is \(F G\) necessarily an antiderivative of \(f g\) ? Explain. d. Is \(F / G\) necessarily an antiderivative of \(f / g ?\) Explain.

Step-by-Step Solution

Answer

(a) Yes; (b) Yes; (c) No; (d) No.

Step 1: Understanding Antiderivatives

An antiderivative of a function is a function whose derivative is the original function. Hence, if \( F \) is an antiderivative of \( f \), then \( F'(x) = f(x) \), and similarly, if \( G \) is an antiderivative of \( g \), then \( G'(x) = g(x) \).

Step 2: Proving (a) - \( F+G \) is an Antiderivative of \( f+g \)

The derivative of the sum of two functions is the sum of their derivatives. Therefore, the derivative of \( F+G \) is \((F+G)'(x) = F'(x) + G'(x) = f(x) + g(x) \). Thus, \( F+G \) is indeed an antiderivative of \( f+g \) on \( I \).

Step 3: Proving (b) - \( c F \) is an Antiderivative of \( c f \)

The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function. Thus, \((c F)'(x) = c F'(x) = c f(x) \). Therefore, \( c F \) is an antiderivative of \( c f \) on \( I \).

Step 4: Evaluating (c) - \( F G \) as an Antiderivative of \( f g \)

The derivative of the product of two functions, \( F \) and \( G \), is \( F'(x)G(x) + F(x)G'(x) \). Applying this, the derivative of \( F G \) would be \( f(x)G(x) + F(x)g(x) \), not \( f(x)g(x) \). Therefore, \( F G \) is not necessarily an antiderivative of \( f g \).

Step 5: Evaluating (d) - \( F / G \) as an Antiderivative of \( f / g \)

Applying the quotient rule, the derivative of \( F/G \) is \( \frac{F'(x)G(x) - F(x)G'(x)}{(G(x))^2} \). This does not simplify to \( f(x)/g(x) \), so \( F/G \) is not necessarily an antiderivative of \( f/g \).

Key Concepts

Sum and Difference of Functions

Antiderivatives have a unique property when it comes to addition and subtraction of functions. If you have two functions, say \( F \) and \( G \), which are antiderivatives of \( f \) and \( g \) respectively, you can easily find the antiderivative of their sum, \( f + g \). This is because the derivative operator follows the rule that the derivative of a sum is the sum of the derivatives.

So, for the functions \( F \) and \( G \), their derivative is given by:

• \((F+G)'(x) = F'(x) + G'(x) = f(x) + g(x)\)

This property ensures that when you combine functions by addition, you can directly add their derivatives. This applies similarly to the difference of functions, where the derivative of \( F - G \) is simply \( F'(x) - G'(x) = f(x) - g(x) \). Hence, \( F + G \) or \( F - G \) will be antiderivatives of \( f+g \) or \( f-g \) respectively.

Constant Multiplication

When you multiply a function by a constant, you can easily determine its antiderivative by understanding the behavior of derivatives with constant multiplication. If \( F \) is an antiderivative of \( f \), then multiplying this antiderivative by a constant \( c \) follows a straightforward rule.

The rule states that:

• \((cF)'(x) = cF'(x) = cf(x)\)

This shows that the derivative of a constant multiple of a function is just the constant multiplied by the derivative of the function. This operation allows \( cF \) to be an antiderivative of \( cf \). Consistent application of this property simplifies calculus operations involving constants.

Derivative of a Product

The process of finding derivatives in products of two functions is more complex than taking the derivative of a sum or a constant multiple. This is due to the product rule, which must be applied.For two functions \( F \) and \( G \), the derivative is calculated by:

• \[ (FG)'(x) = F'(x)G(x) + F(x)G'(x) \]

This formula shows that you need to take the derivative of each function and balance their effects. This means \( FG \) will not necessarily simplify to \( fg \), which indicates that \( FG \) will not always be an antiderivative of \( fg \). This is an important consideration when working with products in calculus, ensuring that you correctly apply the product rule.

Derivative of a Quotient

Finding the derivative of a quotient, where one function is divided by another, follows its own set of rules dictated by the quotient rule. When you're given functions like \( F \) and \( G \), and you need the derivative of their quotient \( F/G \), using the quotient rule becomes necessary.The derivative is given by:

• \[ \left( \frac{F}{G} \right)'(x) = \frac{F'(x)G(x) - F(x)G'(x)}{(G(x))^2} \]

This result shows that simply dividing the derivatives, \( F'(x)/G'(x) \), is not sufficient and doesn't simplify to \( f(x)/g(x) \). Consequently, \( F / G \) doesn't necessarily work as an antiderivative for \( f/g \). Understanding the nuance of the quotient rule is essential for accurately dealing with ratios in calculus problems.