Problem 38 - Calculus - AP Edition

Question

Determine whether the statement is true or false. Explain your answer. $$\text { Find } \frac{d}{d \lambda}\left[\frac{\lambda \lambda_{0}+\lambda^{6}}{2-\lambda_{0}}\right] \quad\left(\lambda_{0} \text { is constant }\right)$$

Step-by-Step Solution

Answer

The statement about the derivative is true; the derivative is \( \frac{\lambda_0 + 6\lambda^5}{2 - \lambda_0} \).

Step 1: Understanding the Problem

We need to determine the derivative of the function \( f(\lambda) = \frac{\lambda \lambda_0 + \lambda^6}{2 - \lambda_0} \) with respect to \( \lambda \). The variable \( \lambda_0 \) is a constant, which means it does not change with \( \lambda \).

Step 2: Identifying the Derivative to Find

The function \( f(\lambda) = \frac{\lambda \lambda_0 + \lambda^6}{2 - \lambda_0} \) is a rational function. The simplest approach is to notice the constant denominator \( 2 - \lambda_0 \), and use the constant multiple rule for derivatives.

Step 3: Applying the Constant Factor Rule

Since \( 2 - \lambda_0 \) is a constant, we can take it out of the derivative, thus:\[\frac{d}{d\lambda} \left[ \frac{\lambda \lambda_0 + \lambda^6}{2 - \lambda_0} \right] = \frac{1}{2 - \lambda_0} \cdot \frac{d}{d\lambda} (\lambda \lambda_0 + \lambda^6)\]

Step 4: Differentiating the Function

Now we differentiate \( \lambda \lambda_0 + \lambda^6 \) with respect to \( \lambda \). The derivative of \( \lambda \lambda_0 \) with respect to \( \lambda \) is \( \lambda_0 \) because \( \lambda_0 \) is a constant. The derivative of \( \lambda^6 \) is \( 6\lambda^5 \). Therefore, \[\frac{d}{d\lambda}(\lambda \lambda_0 + \lambda^6) = \lambda_0 + 6\lambda^5\]

Step 5: Combining the Results

Substitute the derivative back into the expression:\[\frac{d}{d\lambda} \left[ \frac{\lambda \lambda_0 + \lambda^6}{2 - \lambda_0} \right] = \frac{1}{2 - \lambda_0} (\lambda_0 + 6\lambda^5)\]

Step 6: Final Result

Therefore, the derivative is:\[\frac{\lambda_0 + 6\lambda^5}{2 - \lambda_0}\]

Key Concepts

Constant Multiple Rule

When we talk about derivatives, the constant multiple rule is one of the basic rules in calculus that simplifies our work. This rule tells us that if we need to differentiate a function multiplied by a constant, we can "factor out" that constant. This means we can take it outside of the differentiation process.
For example, if you have a function like \( c \cdot f(x) \), where \( c \) is a constant, the derivative of this function can be calculated as \( c \cdot f'(x) \). Instead of differentiating the entire product, you just focus on the function without the constant and multiply the result by the constant.
This rule is useful in situations where constants are involved. Like in our exercise, the denominator \( 2 - \lambda_0 \) is a constant. Using the constant multiple rule allows us to simplify the process of finding the derivative. By taking the constant out, we end up with a simpler differentiation step.

Rational Function

A rational function is one that can be expressed as the quotient of two polynomials. In simple terms, it's a fraction where both the numerator and the denominator are polynomials. For example, \( \frac{p(x)}{q(x)} \) is a rational function where both \( p(x) \) and \( q(x) \) are polynomials.
In the exercise, the function might look complicated at first because it contains a fraction. But understanding it's a rational function means it can be broken down more easily using techniques from calculus, such as differentiation.

• The function \( f(\lambda) = \frac{\lambda \lambda_0 + \lambda^6}{2 - \lambda_0} \) is a clear example, where the top is \( \lambda \lambda_0 + \lambda^6 \) and the bottom is the constant \( 2 - \lambda_0 \).
• Recognizing this structure helps us decide on the most efficient method for differentiation, such as using the constant multiple rule as mentioned before.

Differentiation

Differentiation is a fundamental calculus concept that involves finding the rate at which a function changes at any point. It is like finding how steep a curve is at a particular spot or how quickly something is moving at a specific moment.
When you differentiate a function, you are essentially finding a new function, referred to as the derivative, which tells you the slope of the original function at any given point. Let's consider the function in our example, \( f(\lambda) = \frac{\lambda \lambda_0 + \lambda^6}{2 - \lambda_0} \). To differentiate this:

• First, identify the use of relevant rules, like the constant multiple rule, if applicable.
• Then, apply basic differentiation techniques, focusing here on the sum of \( \lambda \lambda_0 \) and \( \lambda^6 \).

This leads us to finding how each term independently contributes to the overall rate of change. For instance, in this case, the derivatives \( \lambda_0 \) and \( 6\lambda^5 \) were obtained separately before being multiplied by any constant factors.

Power Rule

The power rule is a quick and efficient technique for differentiating functions that involve power terms. It states that if you have a function \( f(x) = x^n \), where \( n \) is any real number, then the derivative \( f'(x) \) is \( n \cdot x^{n-1} \).
This is incredibly powerful because it reduces the time needed to differentiate power terms, which are common in algebraic expressions.

• For instance, in the function \( f(\lambda) = \lambda^6 \), the power rule comes into play. Differentiating using the power rule gives the derivative as \( 6\lambda^{5} \), where the exponent is multiplied by the original coefficient, and the exponent itself is reduced by one.
• Without the power rule, you'd have to rely on more cumbersome methods like the definition of the derivative to achieve the same result.

In our exercise, using the power rule lets us quickly find the derivative of \( \lambda^6 \), simplifying the overall differentiation process greatly.