Problem 32 - Calculus Early Transcendentals

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: Apply the Power Rule to the First Term

The first term is \(2\alpha^{-1}\). Using the power rule \(\frac{d}{dx}x^n = nx^{n-1}\), the derivative of \(\alpha^{-1}\) is \(-\alpha^{-2}\). Therefore, the derivative of \(2\alpha^{-1}\) is \(2 \times -\alpha^{-2} = -2\alpha^{-2}\).

Step 2: Differentiate the Second Term

The second term is \(\alpha\). The derivative of \(\alpha\), which is \(x^1\), is simply \(1\), according to the power rule.

Step 3: Sum the Derivatives

Combine the derivatives obtained from each term: \(-2\alpha^{-2} + 1\). This is the derivative of the entire expression \(\frac{d}{d \alpha}[2 \alpha^{-1} + \alpha]\).

Key Concepts

Using the Power Rule in Calculus

The power rule is a fundamental technique in calculus that is used to find derivatives of functions. This rule is especially useful when dealing with polynomial expressions. The power rule states that if you have a function of the form \(x^n\), the derivative is \(nx^{n-1}\). This means you multiply the variable's exponent by the coefficient, and then subtract one from the exponent.

Let's consider the function \(2\alpha^{-1}\). Applying the power rule here involves:

• Identifying the exponent, which is \(-1\).
• Multiplying the coefficient \(2\) by the exponent \(-1\).
• Subtracting one from the exponent to get \(-2\).

Thus, the derivative of \(2\alpha^{-1}\) becomes \(-2\alpha^{-2}\). This simple method allows us to handle various powers efficiently, whether they are positive, negative, or fractional.

Understanding Differentiation

Differentiation is the mathematical process of finding a derivative, which represents how a function changes as its input changes. Essentially, it measures the rate of change or the slope of the function at any given point.

When we differentiate a function, we apply a set of rules—like the power rule—to find the derivative. In the given exercise, we looked at the expression \(2\alpha^{-1} + \alpha\). Differentiation helped us find the derivative for each part of the expression separately:

• For \(2\alpha^{-1}\), we used the power rule to get \(-2\alpha^{-2}\).
• For \(\alpha\), which is \(\alpha^1\), applying the power rule gives us a simple derivative of \(1\).

By finding these derivatives, we can understand how each component of the expression behaves as \(\alpha\) changes. This gives insight into the overall dynamics of the mathematical function.

Breaking Down Mathematical Expressions

Mathematical expressions often consist of multiple terms that each need to be evaluated to understand the entire function. When differentiating these expressions, it's crucial to approach them term by term.

In our original exercise, the expression \(2\alpha^{-1} + \alpha\) comprises two separate terms, each of which is addressed independently:

• The term \(2\alpha^{-1}\) is assessed using the power rule, resulting in \(-2\alpha^{-2}\).
• The term \(\alpha\) simplifies easily to a derivative of \(1\).

Once we differentiate each term, the next step is to combine these derivatives to form the derivative of the entire expression: \(-2\alpha^{-2} + 1\). This methodical approach ensures each part is handled correctly, leading to an accurate overall result.