Problem 13 - Thomas Calculus

Question

In Exercises \(13-16,\) find \(y^{\prime}(a)\) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate. \(y=\left(3-x^{2}\right)\left(x^{3}-x+1\right)\)

Step-by-Step Solution

Answer

The derivative is \( y' = -5x^4 + 10x^2 - 2x - 3 \).

Step 1: Identify Functions for Product Rule

Given the function \( y = (3-x^2)(x^3-x+1) \), identify the two functions involved in the product. Here, let \( u = 3 - x^2 \) and \( v = x^3 - x + 1 \).

Step 2: Differentiate Functions Individually

First, differentiate the functions individually with respect to \( x \). For \( u = 3 - x^2 \), the derivative is \( u' = -2x \). For \( v = x^3 - x + 1 \), the derivative is \( v' = 3x^2 - 1 \).

Step 3: Apply the Product Rule

Use the product rule for differentiation: \( y' = u'v + uv' \). Substitute the values: \( y' = (-2x)(x^3 - x + 1) + (3 - x^2)(3x^2 - 1) \).

Step 4: Expand the Derivative Expression

Distribute each term: \[ y' = -2x(x^3 - x + 1) + (3 - x^2)(3x^2 - 1) \].Expand both: \( y' = -2x^4 + 2x^2 - 2x + 9x^2 - 3 - 3x^4 - x^2 \).

Step 5: Combine Like Terms

Combine the like terms to simplify the expression: \( y' = -5x^4 +10x^2 - 2x - 3 \).

Step 6: Distribute Product to Expand

Now, multiply the original expressions to expand into simpler terms: \( y = (3-x^2)(x^3-x+1) \).Expanding gives: \( y = 3x^3 - 3x + 3 - x^5 + x^3 - x^2 \).

Step 7: Simplify the Expression

Combine like terms in the expanded expression: \( y = -x^5 + 4x^3 - x^2 - 3x + 3 \).

Step 8: Differentiate the Simplified Expression

Differentiate each term individually: \( y' = -5x^4 + 12x^2 - 2x - 3 \).

Step 9: Compare and Confirm Results

Both methods yield the result: \( y' = -5x^4 + 10x^2 - 2x - 3 \). Thus, both approaches find the derivative accurately.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, which involves finding the rate at which a function changes. When we say we are differentiating a function, we mean that we're calculating its derivative. This is useful when analyzing how functions behave, especially when considering their turning points or slopes at specific points on their graphs. In our exercise, the task is to find the derivative of a polynomial function. By applying the Product Rule, we can take derivatives of functions that are products of two or more functions.
The Product Rule states that if we have a function defined as the product of two functions, say \( u(x) \) and \( v(x) \), then the derivative \( y'(x) \) is given by:

• \( y' = u'v + uv' \)

Where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively. For our polynomial expression \( y = (3-x^2)(x^3-x+1) \), differentiating each part lets us find the overall rate of change in a structured way. This process is crucial in tackling problems that involve more complex functions as well.

Polynomial Functions

Polynomial functions are expressions involving variables raised to various powers, and these variables combine in a linear manner after multiplication by coefficients. Polynomial functions are included within equations such as \( y = ax^n + bx^{n-1} + ... + zx^0 \), where each term represents a part of the polynomial function.
In our exercise, the polynomial function \( y = (3-x^2)(x^3-x+1) \) involves multiplying two polynomial expressions. By using the Product Rule for differentiation, we start by treating each expression as its own function, differentiating them separately before combining the results. However, another approach to handle polynomial differentiation is to simplify the equation into fewer, simpler polynomial terms.
After expansion, for example, we rewrote the function as \( y = -x^5 + 4x^3 - x^2 - 3x + 3 \), which then allows easy differentiation using basic rules, like getting the exponent's coefficient multiplied by the power and reducing the power by one for each term. This approach often simplifies the entire differentiation process.

Chain Rule

The Chain Rule is another differentiation tool essential when dealing with compositions of functions, though not directly applied in our exercise. Learning it is beneficial since functions can often be expressions within expressions. It helps us when we have a function of a function, written as \( f(g(x)) \). In such cases, the Chain Rule states:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

While our specific exercise used the Product Rule, understanding the Chain Rule equips us with another powerful tool for function differentiation. Knowing when to apply each rule is critical, with the Product Rule useful for products of functions, while the Chain Rule is for compositions. Mastering these concepts provides flexibility and precision when tackling more intricate differentiation scenarios in calculus.