Problem 32 - Calculus - AP Edition

Question

Find the indicated derivative. $$\frac{d}{d \alpha}\left[2 \alpha^{-1}+\alpha\right]$$

Step-by-Step Solution

Answer

The derivative is \(-2 \alpha^{-2} + 1\).

Step 1: Understand the Function

The given function is a combination of two terms: \(2 \alpha^{-1}\) and \(\alpha\). We are asked to differentiate this function with respect to \(\alpha\).

Step 2: Differentiate Term by Term

Using the differentiation rules, - For \(2 \alpha^{-1}\), rewrite it as \(2 \alpha^{-1} = 2 \alpha^{-1} = 2 \alpha^{-1}\) which means we use the power rule on \(\alpha^{-1}\).- For \(\alpha\), use the power rule directly.

Step 3: Apply Power Rule to \(2 \alpha^{-1}\)

The power rule states \(\frac{d}{dx}x^n = nx^{n-1}\). For the term \(2 \alpha^{-1}\), this gives:\[\frac{d}{d\alpha}(2 \alpha^{-1}) = 2 \times (-1) \alpha^{-2} = -2 \alpha^{-2}\]

Step 4: Differentiate \(\alpha\)

Since \(\alpha = \alpha^1\), the power rule gives:\[\frac{d}{d\alpha}(\alpha) = 1 \times \alpha^{1-1} = 1 \times \alpha^0 = 1\]

Step 5: Combine the Results

Adding the derivatives of each term together:\[\frac{d}{d\alpha}\left[2 \alpha^{-1} + \alpha\right] = -2 \alpha^{-2} + 1\]

Step 6: Write Final Derivative

The derivative of the given function is:\[\frac{d}{d\alpha}(2 \alpha^{-1} + \alpha) = -2 \alpha^{-2} + 1\]

Key Concepts

Differentiation Rules

Differentiation in calculus is like finding the speed of an object from its position data. The process involves differentiating functions to determine their derivative, which is the rate of change of the function with respect to a variable. In simple terms, it's a way to figure out how fast something is changing at any given point. Differentiation rules provide a systematic approach to finding these derivatives efficiently.
There are several differentiation rules, such as:

• The Power Rule
• Constant Multiple Rule
• Sum Rule
• Product Rule

These rules help make the process of differentiation straightforward by providing shortcuts to calculate derivatives instead of using the definition of derivatives each time. For instance, the Power Rule helps to quickly differentiate polynomial terms. By following these rules, one can determine the rate of change for various types of functions, even when they are complex or involve multiple terms.

Power Rule

The Power Rule is a fundamental differentiation rule that simplifies finding the derivative of polynomial terms. It states that for any term of the form \(x^n\), the derivative can be found using the formula: \(\frac{d}{dx}x^n = nx^{n-1}\).
Let's break this down step by step:

• Identify the exponent \(n\) in the term \(x^n\).
• Multiply the whole term by the exponent \(n\).
• Reduce the exponent by one to get \(nx^{n-1}\).

For example, when differentiating \(\alpha^{-1}\), we apply the power rule:

• Here, \(n = -1\).
• So the derivative is \(-1\cdot\alpha^{-2} = -\alpha^{-2}\).

The Power Rule is useful because it drastically simplifies the calculation of derivatives for polynomial and power functions, saving time and effort.

Term by Term Differentiation

Term by term differentiation is a technique used when functions consist of multiple separate terms. Instead of finding the overall derivative in one go, each term is handled separately. This approach makes calculations easier, especially when dealing with complex expressions.
Here's how it works in practice:

• Identify each separate term in the function. For example, in \(2 \alpha^{-1} + \alpha\).
• Find the derivative of each term individually. Use the appropriate differentiation rules, such as the Power Rule, for each term.
• Combine the individual derivatives into the final derivative of the entire function.

In the exercise we looked at, we differentiated \(2 \alpha^{-1}\) and \(\alpha\) separately. First, we found that the derivative of \(2 \alpha^{-1}\) is \(-2 \alpha^{-2}\). Then, using the Power Rule, we found that the derivative of \(\alpha\) is 1. After calculating derivatives for each term, we simply combined them: \(-2 \alpha^{-2} + 1\). This process illustrates the powerful simplicity of term by term differentiation.