Problem 38 - Calculus Early Transcendentals

Question

Find $$ \frac{d}{d \lambda}\left[\frac{\lambda \lambda_{0}+\lambda^{6}}{2-\lambda_{0}}\right] \quad\left(\lambda_{0} \text { is constant }\right) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{\lambda_{0} + 6 \lambda^5}{2 - \lambda_{0}} \).

Step 1: Identify the Function

The given expression is \( \frac{\lambda \lambda_{0} + \lambda^{6}}{2 - \lambda_{0}} \). We need to find its derivative with respect to \( \lambda \).

Step 2: Simplify the Expression

Since \( \lambda_{0} \) is a constant, the expression can be considered as a sum of two separate terms: \( \frac{\lambda \lambda_{0}}{2 - \lambda_{0}} + \frac{\lambda^{6}}{2 - \lambda_{0}} \).

Step 3: Differentiate the First Term

The first term \( \frac{\lambda \lambda_{0}}{2 - \lambda_{0}} \) is a linear function of \( \lambda \) and can be differentiated as \( \frac{\lambda_{0}}{2 - \lambda_{0}} \).

Step 4: Differentiate the Second Term

The second term \( \frac{\lambda^6}{2 - \lambda_{0}} \) involves the power of \( \lambda \). Use the power rule for differentiation to get \( \frac{6 \lambda^5}{2 - \lambda_{0}} \).

Step 5: Combine the Derivatives

Now, combine the derivatives of the two terms: \( \frac{\lambda_{0}}{2 - \lambda_{0}} + \frac{6 \lambda^5}{2 - \lambda_{0}} \).

Step 6: Simplify the Final Expression

The final derivative is simplified by combining the fractions: \( \frac{\lambda_{0} + 6 \lambda^5}{2 - \lambda_{0}} \).

Key Concepts

Power Rule

When tackling calculus problems, the power rule is an essential tool for finding derivatives of polynomials. It's a quick method to calculate the derivative of any term of the form \( ax^n \), where \( a \) is a constant, \( x \) is the variable, and \( n \) is a real number.

• This powerful rule states that the derivative of \( ax^n \) is \( anx^{n-1} \).
• It's particularly helpful when differentiating terms where the variable \( x \) is raised to a power, as it allows you to lower the power by one and multiply by the original power.

In our exercise, this rule has been applied to differentiate the term \( \frac{\lambda^6}{2 - \lambda_{0}} \). Here, \( n \) is 6, so the derivative becomes \( 6\lambda^5 \). The constant \( \frac{1}{2 - \lambda_{0}} \) remains, factoring into the derivative seamlessly.

Differentiation

Differentiation is the process of finding the derivative of a function. It measures how a function's output changes as inputs vary. This concept is foundational in calculus and is used to solve problems involving rates of change or slopes of curves.

• In practical terms, differentiating an expression helps determine the rate at which one variable changes relative to another.
• The goal of differentiation is to obtain a formula that expresses this rate of change.

In the given exercise, we find the derivative of the function \( \frac{\lambda \lambda_{0} + \lambda^6}{2 - \lambda_{0}} \) by differentiating each term with respect to \( \lambda \). By applying differentiation correctly, the change in the function based on variations in \( \lambda \) becomes clear and manageable.

Algebraic Manipulation

Algebraic manipulation refers to the process of transforming mathematical expressions into equivalent forms to simplify them, making calculations easier to handle. This technique is crucial in differentiation as it often involves simplifying complex expressions before proceeding with differentiation.

• In the context of the exercise, we first separated the given function into two fractions, both with the common denominator \( 2 - \lambda_{0} \).
• This simplification allowed us to differentiate each term separately, leveraging easier algebraic forms.

The final step in finding the derivative included combining the differentiated terms using a common denominator, resulting in a neat and simplified answer: \( \frac{\lambda_{0} + 6 \lambda^5}{2 - \lambda_{0}} \). This highlights the necessity of applying strategic algebraic manipulation alongside calculus rules to achieve a clear solution.