Problem 8 - College Algebra and Calculus: An Applied Approach

Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$ g(x)=\left(x^{2}-2 x+1\right)\left(x^{3}-1\right) $$

Step-by-Step Solution

Answer

Based on the Product Rule, the derivative of the provided function is \(5x^4~-~8x^3~+~3x^2~-~2x~+~2\).

Step 1: Identifying the functions

The given function can be written as a product of two smaller functions. We denote these as \(f(x) = x^2~-~2x~+~1\) and \(h(x) = x^3~-~1\).

Step 2: Applying the Product Rule

The Product Rule is stated as follows: if we have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is given by \(u'(x)v(x) ~+~ v'(x)u(x)\). By applying this rule to our chosen functions, we calculate each derivative to find \(f'(x) = 2x~-~2\) and \(h'(x) = 3x^2\).

Step 3: Calculating the derivative

Following the Product Rule, the derivative of \(g(x)\) is given by \(f'(x)h(x) ~+~ h'(x)f(x) = (2x~-~2)(x^3~-~1) ~+~ (3x^2)(x^2~-~2x~+~1)\).

Step 4: Simplifying the derivative

The derivative can be further simplified to make the math easier to manage. This becomes: \(2x^4~-~2x ~-~ 2x^3 ~+~2 ~+~ 3x^4~-~6x^3~+~3x^2 = 5x^4~-~8x^3~+~3x^2~-~2x~+~2\)

Step 5: Substitute the given x-value

At this point, they could plug in any given x-value into the derived function to find the slope of the original function, \(g(x)\), at that point. However, no specific x-value was given in this exercise.

Key Concepts

Derivative

In calculus, the derivative is a fundamental concept that measures how a function changes as its input changes. It represents the rate at which the function's output is changing at any given point. Understanding derivatives is essential for analyzing functions because they provide insights into the behavior and trends of the function.
To find the derivative of a function, differentiation is used. This process involves determining the instantaneous rate of change or the slope of the tangent line to the function at any point.
Derivatives are not only about calculating slopes but are also employed to understand characteristics like:

• Increases or decreases in functions
• Identifying maximum or minimum points
• Understanding concavity and points of inflection

Thus, mastering the derivative concept allows for a deeper grasp of a function's overall behavior.

Differentiation Rules

Differentiation rules are formulas and principles that simplify finding derivatives of functions. These rules are crucial time-savers when working with complex functions, enabling easier computation and understanding.
In the example provided, the Product Rule is specifically used. This rule simplifies the differentiation of functions that are the product of two or more terms. The Product Rule states:

\[ (u(x) \, v(x))' = u'(x) v(x) + v'(x) u(x) \]

This expression demonstrates how to differentiate the product of two functions by computing the derivatives of the individual functions separately and recombining them.

There are several other differentiation rules like:

• Sum Rule
• Quotient Rule
• Chain Rule

Each rule applies to specific scenarios, helping simplify complex differentiation tasks efficiently. Understanding these can boost your ability to approach calculus problems with confidence.

Function Simplification

Function simplification involves reducing a complex expression into a simpler, more manageable form. It is a useful tool in calculus to make equations easier to handle and interpret.
After finding the derivative using differentiation rules, expressions are often cumbersome. Simplifying these results can:

• Make them more concise
• Provide clearer insights into behavior
• Facilitate easier substitution of specific values

In the original exercise, after applying the Product Rule, the resulting expression was simplified from several extended terms into a cleaner, consolidated form:

\[ 5x^4 - 8x^3 + 3x^2 - 2x + 2 \]

Such simplification steps help in quickly evaluating the function's characteristics without complex arithmetic clutter.