Problem 32 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=\frac{1}{\sqrt[3]{\left(x^{2}+x-9\right)^{2}}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -\frac{2(2x + 1)}{3(x^2 + x - 9)^{5/3}} \).

Step 1: Rewrite the Function

The function is given as \( f(x)=\frac{1}{\sqrt[3]{(x^2+x-9)^{2}}} \). To simplify, rewrite it using exponent rules: \( f(x)=(x^2+x-9)^{-2/3} \). This makes it easier to apply the power rule.

Step 2: Apply the Generalized Power Rule

The Generalized Power Rule states that if \( f(x) = (g(x))^n \), then \( f'(x) = n \cdot g(x)^{n-1} \cdot g'(x) \). Here, \( g(x) = x^2 + x - 9 \) and \( n = -\frac{2}{3} \).

Step 3: Find the Derivative of the Inner Function

Differentiate \( g(x) = x^2 + x - 9 \) to find \( g'(x) \). The derivative is \( g'(x) = 2x + 1 \).

Step 4: Differentiate Using the Power Rule

Substitute into the generalized power rule. We find \( f'(x) = -\frac{2}{3} \cdot (x^2 + x - 9)^{-\frac{5}{3}} \cdot (2x + 1) \).

Step 5: Simplify the Derivative Expression

Simplify the expression by combining terms, so the derivative is \( f'(x) = -\frac{2(2x + 1)}{3(x^2 + x - 9)^{5/3}} \).

Key Concepts

Understanding Derivatives

A derivative represents the rate at which a function changes. When we talk about finding a derivative, we're essentially asking: how does the output of a function change with a small change in input? This is especially useful for understanding trends like speed (how position changes over time) or growth rates in economics.

Finding the derivative involves a process called differentiation. In simple terms, it's finding a slope of the curve at any point. Think of the derivative as a formula that gives you the slope of a function at any specific point. This slope gives us valuable insights into the behavior of functions.

In the exercise, the function is first rewritten to a form that easily applies derivative rules. We convert the given problem into a power of a function using exponent rules. This prepares the function for applying the Generalized Power Rule for derivatives.

Applying the Chain Rule

The Chain Rule is an essential tool in calculus for dealing with compositions of functions. If you have a function within another function, the chain rule helps you differentiate them effectively. It sounds complicated, but think of it as working with layers, where you peel each layer back one by one.

Here's how it works in simple terms:

• If you have two functions combined, you differentiate the outer function first.
• Then, multiply that by the derivative of the inner function.

This is exactly what we do in the problem using the Chain Rule.
In this exercise:

• The outer function is a power of the inner function, rewritten for ease.
• The inner function is the polynomial within this power.
• We find the derivative of this inner part and multiply it by the derivative of the outer part, achieved by the Generalized Power Rule.

This systematic approach allows us to handle even complex functions by breaking them down into manageable pieces.

Mastering Exponent Rules

Exponent rules simplify working with powers of numbers, which is a fundamental skill in calculus. They allow us to rewrite functions in a format ready for differentiation and are crucial for using the Generalized Power Rule effectively.

Key exponent rules used include:

• Converting roots into fractional exponents, such as turning \({\sqrt[3]{a}} \) into \( a^{1/3} \).
• Handling negative exponents, which flips a term between the numerator and denominator.

In our problem:

• The given function involves a cube root in the denominator, rewritten as a power with a negative exponent, making it applicable for calculus tools.

By rewriting functions with exponent rules, we maintain clarity and ensure we can apply the proper calculus techniques efficiently. This step transforms a complex-looking problem into one that closely follows a predictable pattern in calculus exercises.