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Chapter 3: Problem 13

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. \(y=\left(3-x^{2}\right)\left(x^{3}-x+1\right)\)

Short Answer

Expert verified
The derivative is \(y' = -5x^4 + 12x^2 - 2x - 3\).

Step by step solution

01

Identify the Functions

Identify the two functions involved in the product separately. First, let \( u = 3 - x^2 \) and \( v = x^3 - x + 1 \). We will use these in the Product Rule.
02

Use the Product Rule

The Product Rule states that if \( y = uv \), then \( y' = u'v + uv' \). Find the derivatives of \( u \) and \( v \): \( u' = -2x \) and \( v' = 3x^2 - 1 \). Substitute into the Product Rule: \( y' = (-2x)(x^3 - x + 1) + (3 - x^2)(3x^2 - 1) \).
03

Simplify Each Term

Simplify both terms of the derivative expression: \((-2x)(x^3 - x + 1) = -2x^4 + 2x^2 - 2x \) and \((3 - x^2)(3x^2 - 1) = 9x^2 - 3 - 3x^4 + x^2 \).
04

Combine and Simplify the Expression

Combine and simplify the expressions: \(-2x^4 + 2x^2 - 2x + 9x^2 - 3 - 3x^4 + x^2 \). Combine like terms: \(-5x^4 + 12x^2 - 2x - 3 \). This is the derivative found using the Product Rule.
05

Expand the Original Expression

Expand \(y = (3-x^2)(x^3-x+1)\) using distribution: \(3x^3 - 3x + 3 - x^5 + x^3 - x^2 \). Combine like terms to get \(-x^5 + 4x^3 - x^2 - 3x + 3 \).
06

Differentiate the Expanded Expression

Differentiate the expanded expression: \( y' = -5x^4 + 12x^2 - 2x - 3 \). The expression matches the derivative found using the Product Rule.

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Key Concepts

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

Product Rule
In calculus, the Product Rule is a crucial tool for differentiating functions that are the product of two or more expressions. The basic principle here is if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( y = u(x) \, v(x) \) is given by the formula:\[ y^{\prime} = u^{\prime}v + uv^{\prime}\]This rule allows you to take derivatives of more complex expressions efficiently.
In our exercise, we've identified \( u = 3 - x^2 \) and \( v = x^3 - x + 1 \).
Using the product rule, we computed:
  • Derivatives of \( u \) and \( v \) as \( u^{\prime} = -2x \) and \( v^{\prime} = 3x^2 - 1 \).
  • Substituted into the product rule to find \( y^{\prime} \).
This procedure simplifies what could be a cumbersome task, making differentiation of products straightforward and systematic.
Differentiation
Differentiation is the process of finding the derivative of a function, which measures how the function's value changes as its input changes.
It is foundational in calculus, used predominantly for determining rates of change and finding the slope of a curve at any point.
The derivative of a function \( y = f(x) \) is denoted \( y^{\prime} \) or \( f^{\prime}(x) \).In this context, differentiation involves using rules such as the Power Rule and Product Rule.
For instance, if you have a simple power \( x^n \), the derivative \( nx^{n-1} \) is applied.
In the exercise, we differentiated both terms individually first, to make applying the product rule easier.Differentiation is synonymous with understanding the behavior of functions, such as curves and lines, in a visual or physical sense.
Polynomial Derivatives
Understanding polynomial derivatives is important when dealing with expressions involving polynomials, which are algebraic expressions of finite length constructed from variables and constants.Polynomials appear frequently in mathematics and physics because of their versatility and the relative ease of finding their derivatives.
The simplest rule for differentiating polynomials is the Power Rule, which states that for any term of the form \( ax^n \), the derivative is \( anx^{n-1} \).
In our example, both of the functions involved, \( u \) and \( v \), are polynomials. We broke them down term by term using the Power Rule for differentiation:
  • \( (3 - x^2) \) resulted in \( -2x \).
  • \( (x^3 - x + 1) \) resulted in \( 3x^2 - 1 \).
With polynomial derivatives, the emphasis is on organizing and simplifying your results, ensuring the final expression is as concise and informative as possible.

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Most popular questions from this chapter

Find both \(d y / d x\) (treating \(y\) as a differentiable function of \(x\) ) and \(d x / d y\) (treating \(x\) as a differentiable function of \(y )\) . How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. \begin{equation} x y^{3}+x^{2} y=6 \end{equation}

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in g. By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L .\) $$ \begin{array}{l}{\text { a. With } L \text { held constant and } g \text { as the independent variable, }} \\ {\text { calculate } d T \text { and use it to answer parts (b) and (c). }} \\ {\text { b. If } g \text { increases, will } T \text { increase or decrease? Will a pendulum }} \\ {\text { clock speed up or slow down? Explain. }}\end{array} $$ $$ \begin{array}{l}{\text { c. A clock with a } 100 \text { -cm pendulum is moved from a location }} \\ {\text { where } g=980 \mathrm{cm} / \mathrm{sec}^{2} \text { to a new location. This increases the }} \\ {\text { period by } d T=0.001 \mathrm{sec} . \text { Find } d g \text { and estimate the value of }} \\\ {g \text { at the new location. }}\end{array} $$

If \(L=\sqrt{x^{2}+y^{2}}, d x / d t=-1,\) and \(d y / d t=3,\) find \(d L / d t\) when \(x=5\) and \(y=12 .\)

Distance Let \(x\) and \(y\) be differentiable functions of \(t\) and let \(s=\sqrt{x^{2}+y^{2}}\) be the distance between the points \((x, 0)\) and \((0, y)\) in the \(x y\)-plane. a. How is \(d s / d t\) related to \(d x / d t\) if \(y\) is constant? b. How is \(d s / d t\) related to \(d x / d t\) and \(d y / d t\) if neither \(x\) nor \(y\) is constant? c. How is \(d x / d t\) related to \(d y / d t\) if \(s\) is constant?

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2 $$
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