Problem 12 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\). \(f(x)=\frac{x-2}{x^{4}+x+1}\)

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{-3x^4 - 8x^3 + 2x + 3}{(x^4 + x + 1)^2} \).

Step 1: Identify the function type

The function given, \( f(x) = \frac{x-2}{x^4 + x + 1} \), is a rational function because it is the ratio of two polynomials.

Step 2: Apply the Quotient Rule

The Quotient Rule for differentiation is given by: for functions \( u(x) \) and \( v(x) \), it states that \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, let \( u(x) = x - 2 \) and \( v(x) = x^4 + x + 1 \). Find \( u'(x) \) and \( v'(x) \).

Step 3: Differentiate the numerator

Differentiate \( u(x) = x - 2 \). The derivative \( u'(x) \) is \( 1 \).

Step 4: Differentiate the denominator

Differentiate \( v(x) = x^4 + x + 1 \). The derivative \( v'(x) = 4x^3 + 1 \).

Step 5: Substitute into the Quotient Rule

Substitute \( u'(x) = 1 \), \( u(x) = x - 2 \), \( v(x) = x^4 + x + 1 \), and \( v'(x) = 4x^3 + 1 \) into the Quotient Rule: \[ f'(x) = \frac{(1)(x^4 + x + 1) - (x - 2)(4x^3 + 1)}{(x^4 + x + 1)^2} \]

Step 6: Simplify the expression

Simplify the expression for \( f'(x) \): - Expand \((1)(x^4 + x + 1)\) to \(x^4 + x + 1\).- Expand \((x - 2)(4x^3 + 1)\) to \(4x^4 - 8x^3 + x - 2\). - Combine and simplify to get \[ f'(x) = \frac{-3x^4 - 8x^3 + 2x + 3}{(x^4 + x + 1)^2} \]

Key Concepts

Quotient Rule

Differentiation is a key topic in calculus, and many functions need specific methods for finding their derivatives. One such method is the Quotient Rule. This rule is especially useful when dealing with rational functions, which are essentially ratios of two polynomial functions. Whenever you have a function of the form

• \( g(x) = \frac{u(x)}{v(x)} \)

you'll want to use the Quotient Rule to differentiate it efficiently. The rule itself states:

• \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \)

This formula helps find the derivative of the quotient of two functions by:

• First differentiating the numerator \( u \), resulting in \( u' \).
• Then differentiating the denominator \( v \), resulting in \( v' \).
• Finally, applying the rule’s formula to obtain the derivative of the entire function.

Using the Quotient Rule can simplify what seems to be a complex differentiation task into more manageable steps. This breakdown enables more precise calculations without common mistakes. This concept lies at the core of understanding how to differentiate combinations of functions represented as fractions.

Rational Function

Rational functions play a significant role in both mathematics and real-world applications. These functions are expressed in the form of a fraction, where both the numerator and the denominator are polynomials. An example of such a function is

• \( f(x) = \frac{x-2}{x^4 + x + 1} \)

Identifying a function as rational is crucial for choosing the correct approach when differentiating. With rational functions, the Quotient Rule, as we discussed, is your go-to method. This type of function can describe many behaviors, such as rates, proportions, and other phenomena where one quantity depends on another.

• Introducing asymptotes, behavior at infinity, or finding limits often involves rational functions.
• When you break down the function for differentiation, you examine the polynomial structures on the numerator and denominator separately.

Understanding the nature of rational functions improves both comprehension and problem-solving skills when faced with complex calculus problems. Being able to manipulate and differentiate these functions empowers you to handle various mathematical challenges.

Polynomial

Polynomials form the building blocks of many more advanced algebraic functions. A polynomial function consists of terms in the form \( a_nx^n \) where \( n \) is a non-negative integer, and \( a_n \) are coefficients. For instance, in \( x^4 + x + 1 \), the polynomial consists of powers of \( x \) with their respective coefficients.

• This function has degrees in the descending order: 4, 1, and 0, representing each term.
• Simplifying each component of a polynomial using basic rules is key during differentiation.

Derivative of polynomial terms are usually straightforward: apply the power rule where

• the derivative of \( x^n \) is \( nx^{n-1} \).

Such simplicity in differentiation is why polynomials serve as an excellent starting point in calculus learning. They often feature in the numerator and denominator when dealing with rational functions, reinforcing their importance in a wide array of mathematical operations. Mastery over polynomials can unlock a deeper understanding of calculus as a whole and greatly reduce errors in complex calculations.