Problem 19 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ w(z)=\sqrt[3]{9 z-1} $$

Step-by-Step Solution

Answer

The derivative is \( w'(z) = 3(9z - 1)^{-2/3} \).

Step 1: Rewrite the Function

First, it's helpful to express the function in a form that makes differentiation easier. The original function is \( w(z) = \sqrt[3]{9z - 1} \). Rewrite this using exponent notation: \( w(z) = (9z - 1)^{1/3} \).

Step 2: Apply the Generalized Power Rule

The Generalized Power Rule states that if \( y = (f(x))^n \), then the derivative \( y' \) is given by \( y' = n(f(x))^{n-1}f'(x) \). In our case, \( f(z) = 9z - 1 \) and \( n = \frac{1}{3} \).

Step 3: Differentiate the Inner Function

Differentiate the inner function \( f(z) = 9z - 1 \). The derivative is \( f'(z) = 9 \).

Step 4: Substitute and Differentiate

Apply the Generalized Power Rule: substitute \( n = \frac{1}{3} \), \( f(z) = 9z - 1 \), and \( f'(z) = 9 \) into the rule. Thus, we get: \( w'(z) = \frac{1}{3}(9z - 1)^{1/3 - 1} \times 9 \).

Step 5: Simplify the Expression

Now simplify the expression: first, calculate \( 1/3 - 1 = -2/3 \). Therefore, \( w'(z) = \frac{1}{3} \times 9 \times (9z - 1)^{-2/3} = 3 (9z - 1)^{-2/3} \).

Key Concepts

Understanding Derivatives

A derivative represents the rate of change of a function with respect to one of its variables. In simple terms, it tells us how a function is changing at any given point. When we say we're finding the derivative of a function like \( w(z) \), we're essentially looking for the function that describes the instantaneous rate of change of \( w \) with respect to \( z \). This is essential for understanding behaviors like acceleration in physics, or changes in economics.

• It is often denoted by \( f'(x) \) for a function \( f \).
• Derivatives can show where functions are increasing or decreasing.
• They can also indicate the slope of a function's graph.

Derivatives form the backbone of calculus, making them critical in exploring complex mathematical models.

Exponent Notation Basics

Exponent notation is a way of expressing numbers that makes them easier to work with, especially when dealing with powers and roots. In the context of derivatives, it helps convert root forms into exponential forms, paving the way for simpler differentiation. Here, \( w(z) = \sqrt[3]{9z - 1} \) was rewritten as \( w(z) = (9z - 1)^{1/3} \).

• Exponent notation makes the application of rules like the Generalized Power Rule straightforward.
• It allows mathematicians to handle complex expressions in a more manageable way.
• Understanding this notation is crucial for simplification and solving equations.

Overall, exponent notation transforms problems into more calculus-friendly forms, which is particularly useful in differentiation tasks.

Differentiation Rules in Action

Differentiation rules are the guidelines that help us find the derivative of a function efficiently. The Generalized Power Rule is a prime example and is an extension of the Power Rule, a fundamental rule of differentiation.
The Generalized Power Rule states: if you have a function \( y = (f(x))^n \), its derivative \( y' \) can be found using \( y' = n(f(x))^{n-1}f'(x) \).

• This rule applies to functions raised to any power, including fractions and negative numbers.
• It helps in scenarios where multiple function layers are present, as seen in composite functions.
• By separating the function into its outer and inner parts, each is tackled via differentiation, simplifying complex problems.

Knowing when and how to apply these rules streamlines the differentiation process.

Simplifying Expressions

Once derivatives are calculated, the final expression often needs simplification. This involves reducing the complexity to make it easier to interpret or apply. In the given exercise, after applying the Generalized Power Rule, the expression \( w'(z) = \frac{1}{3}(9z - 1)^{-2/3} \times 9 \) transformed into \( 3(9z - 1)^{-2/3} \).

• Subtract exponents where necessary, as seen in changing \( 1/3 - 1 \) to \( -2/3 \).
• Combine constants to reduce the number of terms for simplicity.
• Rearranging terms helps in presenting the derivative in its neatest form.

Simplification is the final step that distills complex equations into useful mathematical insight, making them easier to use in further calculations.