Problem 34 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=\left[\left(x^{3}+1\right)^{2}-x\right]^{4} $$

Step-by-Step Solution

Answer

Apply the Generalized Power Rule: \( f'(x) = 4[(x^3 + 1)^2 - x]^3 [6x^2(x^3 + 1) - 1] \).

Step 1: Apply the Generalized Power Rule

To differentiate the function \( f(x)=\left[\left(x^{3}+1\right)^{2}-x\right]^{4} \), apply the Generalized Power Rule. This rule states that if \( y = [g(x)]^n \), then \( y' = n[g(x)]^{n-1}g'(x) \). Here, set \( g(x) = (x^3 + 1)^2 - x \) and \( n = 4 \). Then, \( f'(x) = 4\left[(x^3 + 1)^2 - x\right]^{3} \cdot g'(x) \).

Step 2: Find the Derivative of the Inner Function

Next, find \( g'(x) \) where \( g(x) = (x^3 + 1)^2 - x \). Start by applying the power rule to \( h(x) = (x^3 + 1)^2 \), setting \( h(x) = u^2 \) and \( u = x^3 + 1 \). The derivative \( h'(x) \) is \( 2u^{1}u'(x) \). Substitute back \( u = x^3 + 1 \) and \( u'(x) = 3x^2 \) to get \( h'(x) = 2(x^3+1) \cdot 3x^2 = 6x^2(x^3 + 1) \).

Step 3: Combine Derivatives in the Inner Function

The derivative of \( g(x) = (x^3 + 1)^2 - x \) is \( g'(x) = h'(x) - 1 \). From Step 2, substitute \( h'(x) = 6x^2(x^3 + 1) \) into the equation to get \( g'(x) = 6x^2(x^3 + 1) - 1 \).

Step 4: Substitute Back and Simplify

Finalize the solution by substituting \( g'(x) \) into the derivative we found using the Generalized Power Rule. So, \( f'(x) = 4\left[(x^3 + 1)^2 - x\right]^{3} \cdot \left[6x^2(x^3 + 1) - 1\right] \). This is the derivative of \( f(x) \).

Key Concepts

Understanding Derivatives

In calculus, derivatives play a crucial role. Essentially, they measure how a function changes as its input changes. They tell us the slope of the function at any given point. This is like finding how steep a hill is at a specific spot you're standing on.
Derivatives are used to solve various problems in fields like physics, engineering, and economics, as they help understand the behavior of changing quantities.
Specifically, finding a derivative involves differentiation, which breaks down into applying rules like the power, product, and chain rules. Understanding these rules allows us to tackle more complex functions easily.

Diving into Calculus Differentiation

Calculus differentiation involves various methods to find derivatives of functions. It's not just about applying a rule; it's about knowing which rule to use and when.
For the problem at hand, we needed to use the Generalized Power Rule. This rule is a part of the differentiation techniques that simplify the process when dealing with composite functions raised to a power.
To efficiently differentiate functions in calculus, one must:

• Recognize type and structure of the function
• Apply the right differentiation rule(s)
• Break down complex functions into simpler parts

By practicing these methods, complex functions become more manageable, and the understanding of calculus deepens with time.

Exploring Higher-Order Functions

Higher-order functions are functions that operate on other functions. These can either take another function as an argument or return them as results. In calculus, when you find a derivative of a function, you're actually creating a new function that describes the rate of change.
In this exercise, while finding the derivative of a complex function, the result is another function (\(f'(x)\)) which can be further differentiated if needed. This process can continue, leading to second derivatives, third derivatives, and so on.
Understanding higher-order functions in calculus enables students to explore the intricacies of how rates of change not only occur but interact at multiple levels in various dimensions.