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Chapter 5: Problem 19

Perform the indicated multiplications in exercises. $$ D(x D-1) $$

Short Answer

Expert verified
The result is \( D + xD^2 \).

Step by step solution

01

Understand the Expression

The expression given is \( D(xD - 1) \). Here, \( D \) is a differential operator that acts on everything to its right. The expression requires you to distribute \( D \).
02

Distribute the Operator

Apply the distributive property to \( D(xD - 1) \), which means \( D \) will act on each term inside the parenthesis separately. This results in two separate expressions: \( D(xD) - D(1) \).
03

Apply the Differential Operator

For the term \( D(xD) \), note that \( D \) means taking a derivative with respect to \( x \). The derivative of a product like \( xD \) (using \( D \) as a constant on its right side) is \( D + xD^2 \) by the product rule. For \( D(1) \), the result is simply zero since the derivative of a constant is zero.
04

Combine the Results

Now, combine the results from the differentiated expressions: \( D(xD) - D(1) = D + xD^2 + 0 \). Hence, the final result of the original multiplication is \( D + xD^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Derivatives
In calculus, a derivative represents how a function changes as its input changes. Think of it as a tool to measure the rate of change or slope at any given point of a function.
  • Derivatives are fundamental to calculus and are used in numerous fields, such as physics and engineering, to understand how variables change in relation to one another.
  • Mathematically, if you have a function \( f(x) \), the derivative \( f'(x) \) represents the function's rate of change.
The symbol \( D \) here represents the differentiation with respect to \( x \). So, when you see \( D(x D) \), it implies applying the derivative operation to the product \( x D \). Once you take the derivative, you understand how the function changes at each point when \( x \) changes minimally.
Exploring the Product Rule
The product rule is a specific rule used to find the derivative of the product of two or more functions.
  • It states that if you have two functions \( u(x) \) and \( v(x) \), the derivative of their product is \( u'v + uv' \).
  • In simple terms, differentiate the first function, multiply it by the second function, then do the reverse.
When dealing with \( D(xD) \) in the exercise, you apply the product rule. Here, you treat \( x \) as one function and \( D \) as another (consider \( D \) as a constant concerning taking a derivative). The result of applying the product rule is \( D + xD^2 \). This result demonstrates the rate of change for the product \( xD \). It's a systematic way of handling derivatives when multiplication is involved.
Applying the Distributive Property
The distributive property is a foundational principle in algebra that states: to multiply a number by a group of numbers added together, you can multiply each separately and add the results.
  • This property often simplifies expressions and helps solve multiplying expressions more flexibly.
  • In mathematics, it is typically expressed as \( a(b + c) = ab + ac \).
In the exercise, when you have \( D(xD - 1) \), the distributive property helps break down the expression. The differential operator \( D \) is distributed into the terms inside the parentheses: \( xD \) and \(-1\). As a result, you evaluate it as \( D(xD) - D(1) \). This method ensures accurate application of the operator across all parts of the expression.

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Most popular questions from this chapter

Show that \(1, \sin x, \cos x\) are linearly independent.

Perform the indicated multiplications in exercises. $$ (D+x)(D-x) $$

Perform the indicated multiplications in exercises. $$ (D-x)(D+x) $$

Show that the functions \(e^{x}, \cos x, \sin x\) are linearly independent.

Show that \(1, \sin ^{2} x, \cos ^{2} x\) are linearly dependent.
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