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Chapter 1: Problem 69

Differentiate each function. \(f(x)=\frac{(x-1)\left(x^{2}+x+1\right)}{x^{4}-3 x^{3}-5}\)

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{-x^6 + 4x^3 - 24x^2}{(x^4 - 3x^3 - 5)^2} \).

Step by step solution

01

Identify the derivative rule to use

The given function is a fraction or rational function of polynomials. To differentiate it, we'll use the quotient rule which states:\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]where \( u \) is the numerator \( (x-1)(x^2+x+1) \) and \( v \) is the denominator \( x^4-3x^3-5 \).
02

Differentiate the Numerator

First, expand the numerator function \( u=(x-1)(x^2+x+1) \) into a polynomial:\[ u = x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1 \]Now, find the derivative of \( u \):\[ u' = \frac{d}{dx}(x^3 - 1) = 3x^2 \].
03

Differentiate the Denominator

Differentiate the denominator function \( v = x^4 - 3x^3 - 5 \):\[ v' = \frac{d}{dx}(x^4 - 3x^3 - 5) = 4x^3 - 9x^2 \].
04

Apply the Quotient Rule

Substitute \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula:\[ f'(x) = \frac{(3x^2)(x^4 - 3x^3 - 5) - (x^3 - 1)(4x^3 - 9x^2)}{(x^4 - 3x^3 - 5)^2} \].
05

Simplify the Derivative Expression

Expand and simplify the expression. Calculate the terms:1. \( 3x^2(x^4 - 3x^3 - 5) = 3x^6 - 9x^5 - 15x^2 \)2. \( (x^3 - 1)(4x^3 - 9x^2) = 4x^6 - 9x^5 - 4x^3 + 9x^2 \)Now combine:\[ f'(x) = \frac{3x^6 - 9x^5 - 15x^2 - 4x^6 + 9x^5 + 4x^3 - 9x^2}{(x^4 - 3x^3 - 5)^2} \]Simplify the expression:\[ f'(x) = \frac{-x^6 + 4x^3 - 24x^2}{(x^4 - 3x^3 - 5)^2} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of Rational Functions
Rational functions are quotients where both the numerator and the denominator are polynomials. The process of finding their derivatives requires special rules beyond basic differentiation. The quotient rule is specifically designed to deal with such situations. It simplifies the process by providing a structured formula:
  • For a given function \( f(x) = \frac{u}{v} \), the derivative \( f'(x) \) is calculated using the formula \( f'(x) = \frac{u'v - uv'}{v^2} \).
  • Here, \( u \) and \( v \) represent the numerator and the denominator, respectively.
To apply the quotient rule, follow these steps:
  • Differentiating the numerator \( u \) to get \( u' \).
  • Differentiating the denominator \( v \) to get \( v' \).
  • Then substitute \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula.
This technique is vital to properly manage the derivatives of rational functions without mistakenly multiplying or dividing components incorrectly.
Polynomial Differentiation
Polynomials are expressions consisting of variables raised to whole number powers and their constant coefficients. The differentiation of polynomials is foundational, involving rules such as the power rule. The process is fairly straightforward:
  • For a term \( ax^n \), its derivative is found using the power rule: \( \frac{d}{dx}(ax^n) = nax^{n-1} \).
  • Each term in the polynomial is differentiated individually, and then they're combined together again.
When we differentiate the polynomial numerator \( (x-1)(x^2+x+1) \) as seen in our example:
  • First, it is expanded to a single polynomial \( x^3 - 1 \).
  • Then differentiated term by term, here yielding \( 3x^2 \) for the derivative.
Similarly, the denominator \( x^4-3x^3-5 \) is handled by identifying each term and finding the derivatives to get \( 4x^3 - 9x^2 \). Understanding and applying polynomial differentiation is crucial for efficiently dealing with both the numerator and the denominator of rational functions.
Simplifying Derivatives
After applying the quotient rule, simplifying the resulting expression can often be the most challenging yet crucial step. Simplification involves systematically reducing the expression to its simplest form:
  • Begin by carefully expanding products using the distributive property. In our example: \( 3x^2(x^4 - 3x^3 - 5) \) becomes \( 3x^6 - 9x^5 - 15x^2 \).
  • Follow similarly for the other expansion \((x^3 - 1)(4x^3 - 9x^2)\).
  • Combine like terms from these expansions to finalize the numerator.
In our transformed exercise, this leads to combining terms through addition and subtraction, ultimately simplifying the numerator to \(-x^6 + 4x^3 - 24x^2\). Simplification reduces computational complexity and makes the expression more comprehensible. It ensures accuracy and clarity, especially when dealing directly with higher-degree polynomials.

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Most popular questions from this chapter

On the moon, all free-fall distance functions are of the form \(s(t)=0.81 t^{2},\) where \(t\) is in seconds and \(s(t)\) is in meters. An object is dropped from a height of 200 meters above the moon. After \(t=2 \mathrm{sec}\) a) How far has the object fallen? b) How fast is it traveling? c) What is its acceleration? d) Explain the meaning of the second derivative of this free-fall function.

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 5}\left(\frac{x^{2}-25}{2 x-10}\right) $$

The function given by \(R(x)=11.74 x^{1 / 4}\) can be used to approximate the maximum range \(\mathrm{R}(x)\) in miles, of an ARSR-3 surveillance radar with a peak power of \(x\) watts (W). (Source: Introduction to RADAR Techniques, Federal Aviation Administration, \(1988 .\). a) Find the rate at which the maximum radar range changes as peak power increases from \(40,000 \mathrm{~W}\) to \(60,000 \mathrm{~W}\) b) Find \(\frac{R(60,000)-R(50,000)}{60,000-50,000}\). What does this rate represent?

Find \(y^{\prime}\) $$ \text { If } y=\frac{1}{3 x^{4}}, \text { find }\left.\frac{d y}{d x}\right|_{x=-1} $$

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow \infty}\left(\frac{4 x^{2}+x-3}{2 x^{2}+1}\right) $$
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