Problem 56 - Calculus Early Transcendentals

Question

Find the indicated derivative. $$ \lambda=\left(\frac{a u+b}{c u+d}\right)^{6} ; \text { find } \frac{d \lambda}{d u} \quad(a, b, c, d \text { constants }) $$

Step-by-Step Solution

Answer

\( \frac{d \lambda}{du} = 6(ad-bc) \frac{(au+b)^5}{(cu+d)^7} \).

Step 1: Recognize the Outer Function

Notice that the expression \( \lambda = \left( \frac{au+b}{cu+d} \right)^6 \) is a power function. The outer function here is \( x^6 \) where \( x = \frac{au+b}{cu+d} \). We will use the chain rule to find the derivative.

Step 2: Apply the Chain Rule

According to the chain rule, the derivative of \( \lambda \) with respect to \( u \) is \( 6 \left( \frac{au+b}{cu+d} \right)^5 \) times the derivative of the inner function \( \frac{au+b}{cu+d} \) with respect to \( u \).

Step 3: Differentiate the Inner Function

To find the derivative of \( \frac{au+b}{cu+d} \) with respect to \( u \), use the quotient rule: \[ \frac{d}{du} \left( \frac{au+b}{cu+d} \right) = \frac{(cu+d) \cdot a - (au+b) \cdot c}{(cu+d)^2} \].

Step 4: Simplify the Derivative of the Inner Function

Simplifying the derivative from the previous step gives: \[ \frac{(cu+d) \cdot a - (au+b) \cdot c}{(cu+d)^2} = \frac{ac(u) + ad - ac(u) - bc}{(cu+d)^2} = \frac{ad-bc}{(cu+d)^2} \].

Step 5: Combine Results

Plug the simplified result into the chain rule expression: \[ \frac{d \lambda}{du} = 6 \left( \frac{au+b}{cu+d} \right)^5 \cdot \frac{ad-bc}{(cu+d)^2} \].

Step 6: Final Expression

The expression can be rewritten as: \[ \frac{d \lambda}{du} = 6(ad-bc) \frac{(au+b)^5}{(cu+d)^7} \] by combining the powers of \((cu+d)\) in the denominator.

Key Concepts

Chain Rule in Derivative Calculus

The chain rule is a fundamental technique in calculus, especially useful when dealing with composite functions. Understanding it is crucial to mastering derivatives.

The chain rule helps us differentiate functions of functions, or composite functions. For instance, if you have a function inside another, like \((f(g(x)))\), the chain rule provides a strategy for finding its derivative. The method involves taking the derivative of the outer function, and multiplying it by the derivative of the inner function.

In the given problem, the expression \( \lambda = \left( \frac{au+b}{cu+d} \right)^6 \) is composite. The outer function is the power \(x^6\) and the inner function is the fraction \(\frac{au+b}{cu+d}\). To apply the chain rule here:

• First, differentiate the outer function: the derivative of \((x^6)\) is \(6x^5\).
• Second, multiply by the derivative of the inner function \(\frac{au+b}{cu+d}\), which we find using another rule – the quotient rule.

By chaining these derivatives together, we efficiently solve for the derivative of complex expressions.

Quotient Rule for Differentiation

The quotient rule is imperative when differentiating a fraction of two functions. In mathematics, this rule states that the derivative of a quotient \(\frac{f(x)}{g(x)}\) is given by:\[\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}\]

For our original exercise, the quotient \(\frac{au+b}{cu+d}\) fits exactly into this rule. The process is like this:

• The numerator \(f(u) = au+b\) and its derivative \(f'(u) = a\).
• The denominator \(g(u) = cu+d\) and its derivative \(g'(u) = c\).
• Apply the quotient rule:\[\frac{d}{du} \left( \frac{au+b}{cu+d} \right) = \frac{(cu+d)(a) - (au+b)(c)}{(cu+d)^2}\]

Simplification is often needed to get a usable form, simplifying gives the clean result \(\frac{ad-bc}{(cu+d)^2}\). Knowing how and when to apply the quotient rule is fundamental when differentiating rational expressions.

Differentiation and Its Application

Differentiation is the process in calculus used to find the rate at which a quantity changes. It's the core concept for understanding derivatives. When you differentiate a function, you're finding another function that represents the slope of the original function at any point.

For the exercise at hand, differentiation is used to find \(\frac{d \lambda}{du}\), the derivative of \(\lambda\) with respect to \(u\). In layman's terms, it tells us how \(\lambda\) changes as \(u\) changes. We do this by differentiating the entire function, using tools like the chain rule and quotient rule, to break down complex forms and find slopes or rates of change.

• Start with recognizing the type of function.
• Apply rules such as the chain rule for composite functions and the quotient rule for rational expressions.
• Combine results into a finalized expression, simplifying when necessary to get a clearer view of the derivative.

In our example, this process led to a neatly simplified expression that shows how all parts of the function contribute to its overall rate of change. Mastery of differentiation techniques is essential in calculus, providing clarity and foresight into dynamic systems.