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Chapter 6: Problem 6

Is there an example of two functions, \(u(x)\) and \(v(x),\) for which \([u(x) \times v(x)]^{\prime}=u^{\prime}(x) \times v^{\prime}(x) ?\)

Short Answer

Expert verified
Yes, for constant functions like \(u(x) = c\) and \(v(x) = c\), this condition is satisfied.

Step by step solution

01

Interpreting the Derivative Condition

We are given the condition \([u(x) \times v(x)]^{\prime} = u^{\prime}(x) \times v^{\prime}(x)\). According to the product rule, \([u(x) \times v(x)]^{\prime} = u^{\prime}(x) v(x) + u(x) v^{\prime}(x)\). We need to check if this matches with \(u^{\prime}(x) \times v^{\prime}(x)\).
02

Testing the Zero Function Example

Consider \(u(x) = c\) and \(v(x) = c\), where \(c\) is a constant function. Then \(u^{\prime}(x) = 0\) and \(v^{\prime}(x) = 0\). So, \([u(x) \times v(x)]^{\prime} = (c \times c)^{\prime} = 0\), and \(u^{\prime}(x) \times v^{\prime}(x) = 0 \times 0 = 0\). They are equal, meeting the condition.
03

Verifying Another Possible Case

If both functions are constant, for instance, choose \(u(x) = 1\) and \(v(x) = 1\). Then, \(u^{\prime}(x) = 0\) and \(v^{\prime}(x) = 0\). Hence, \([u(x) \times v(x)]^{\prime} = (1 \times 1)^{\prime} = 0\), and \(u^{\prime}(x) \times v^{\prime}(x) = 0 \times 0 = 0\). Therefore, the condition holds true.

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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 an essential tool in calculus used when differentiating two functions that are multiplied together. It helps to find the derivative of a product of functions. Imagine you have two functions, \(u(x)\) and \(v(x)\), and they are combined as \(u(x) \times v(x)\). Whereas computing the derivative of each function separately is straightforward, doing so for their product requires the product rule. It states that:
  • \([u(x) \times v(x)]' = u'(x) v(x) + u(x) v'(x)\)
This formula might look complex, but it mirrors a fundamental dynamic: the changing nature of each component contributes to the rate of change of the product. If either function changes, the product itself will change, capturing how these changes interact.
The product rule is particularly valuable because it allows for the differentiation of products without needing to expand the product first. This is especially helpful when you have complex functions, allowing you to sidestep extensive algebraic manipulations.
Understanding Differentiation
Differentiation refers to the process of finding the derivative of a function, which effectively measures how a function changes as its input changes. The derivative is like a rate of change or the slope of the function at any given point. Imagine driving a car: the speed at which you move is the derivative of your position relative to time.
The standard notation for derivatives involves using a prime symbol like \(u'(x)\) or \(v'(x)\). To differentiate a function means to apply specific rules, such as the product rule, chain rule, or quotient rule, depending on the situation.
  • The power rule, for example, is for differentiating simple polynomials.
  • The product rule is used for multiplying functions.
  • The chain rule is applied for composite functions.

For a consistent check on learning, remember that differentiation transforms functions into their derivative form, yielding valuable insights into their behavior. Differentiation helps determine slopes, analyze growth trends, and solve problems involving rates, among other applications in mathematics and science.
Exploring Functions and Their Properties
Functions represent the relationship between two variables, usually denoted as \(x\) (the independent variable) and \(f(x)\) or \(y\) (the dependent variable). Functions can take many forms 鈥� they may be linear, quadratic, exponential, or something more complex. Knowing the type and properties of a function, such as continuity or differentiability, is fundamental in calculus.
Functions often have unique characteristics that define their graphs, behavior, and applications. Consider these basics:
  • Linear functions have a constant rate of change and form straight lines.
  • Quadratic functions create parabolas, impacting various fields like physics with projectile motion.
  • Exponential functions rapidly increase or decrease, critical in studies of growth and decay.
Understanding the behavior of functions is pivotal for correctly applying differentiation rules like the product rule. When solving calculus problems involving multiple functions, recognizing how they work individuall and together ensures precise and effective solutions.

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Most popular questions from this chapter

Let \(P(t)=u(t) / v(t)\). Then $$ \ln P(t)=\ln \left(\frac{u(t)}{v(t)}\right)=\ln u(t)-\ln v(t) $$ Compute the derivative of the two sides of Equation 6.12 using the logarithm chain rule and show that $$ P^{\prime}(t)=\frac{u(t) v^{\prime}(t)-u^{\prime}(t) v(t)}{v^{2}(t)} $$

The word differentiate means 'find the derivative of'. Differentiate a. \(P(t)=\frac{e^{-3 t}}{t^{2}}\) d. \(P(t)=t^{2} e^{2 t}\) g. \(P(t)=\left(e^{t}\right)^{5}\) b. \(P(t)=e^{2 \ln t}\) e. \(P(t)=e^{t \ln 2}\) h. \(P(t)=\frac{t-1}{t+1}\) c. \(P(t)=e^{t} \ln t\) f. \(P(t)=e^{2}\) i. \(P(t)=\frac{3 t^{2}-2 t-1}{t^{-} 1}\)

Exercise 6.2 .4 The Doppler effect. You are standing 100 meters south of a straight train track on which a train is traveling from west to east at the speed 30 meters per second. See Figure 6.2 .4 . Let \(y(t)\) be the distance from the train to you and \(|x(t)|\) be the distance from the train to the point on the track nearest you; \(x(t)\) is negative when the train is west of the point on the track nearest you. a. Write \(y(t)\) in terms of \(x(t)\). b. Find \(y^{\prime}(t)\) for a time \(t\) at which \(x(t)=-200\) c. Time is measured so that \(x(0)=0 .\) Write an equation for \(x(t)\). d. Write an equation for \(y^{\prime}(t)\) in terms of \(t\). e. The whistle from the train projects sound waves at frequency \(f\) cycles per second. The frequency, \(f_{L},\) of the sound reaching your ear is $$ f_{L}(t)=\frac{331.4}{331.4+y^{\prime}(t)} f \frac{\text { cycles }}{\text { second }} $$ \(331.4 \mathrm{~m} / \mathrm{s}\) is the speed of sound in air. Draw a graph of \(f_{L}(t)\) assuming \(f=500\). Derivation of Equation 6.18 for the Doppler effect. A sound of frequency \(f\) traveling in still air has wave length \((331.4 \mathrm{~m} / \mathrm{s}) /(f\) cycles \(/ \mathrm{s})=(331.4 / \mathrm{f}) \mathrm{m} /\) cycle. If the source of the sound is moving at a velocity \(v\) with respect to a listener, the wave length of the sound reaching the listener is \(((331.4+v) / f) \mathrm{m} /\) cycle. These waves travel at \(331.4 \mathrm{~m} / \mathrm{sec},\) and the frequency \(f_{L}\) of these waves reaching the listener is $$ f_{L}=\frac{331.4 \mathrm{~m} / \mathrm{s}}{((331.4+v) / f) \mathrm{m} / \text { cycle }}=\frac{331.4}{331.4+v} f \frac{\text { cycles }}{\text { second }} $$ High frequency sound waves may be used to measure the rate of blood flow in an artery. A high frequency sound is introduced on the skin surface above the artery, and the frequency of the waves reflected from the arterial flow is measured. The difference in frequencies emitted and received is used to measure blood velocity. A train and track with listener location. As drawn, \(x(t)\) is negative

Use the chain rule to differentiate \(P(t)\) for a. \(P(t)=e^{\left(-t^{2}\right)}\) b. \(P(t)=\left(e^{t}\right)^{2}\) c. \(P(t)=e^{2 \ln t}\) d. \(P(t)=\ln e^{2 t}\) e. \(P(t)=\ln (2 \sqrt{t})\) f. \(P(t)=\sqrt{2 \ln t}\) g. \(P(t)=\sqrt{e^{2 t}}\) h. \(P(t)=\sqrt{e^{\left(-t^{2}\right)}}\) i. \(\quad P(t)=\left(t+e^{-2 t}\right)^{4}\) j. \(P(t)=\left(1+e^{\left(-t^{2}\right)}\right)^{-1}\) k. \(P(t)=\frac{3}{4}\left(1-x^{2} / 16\right)^{1 / 2}\) 1\. \(P(t)=(t+\ln (1+2 t))^{2}\)

A useful special case of the quotient formula is the reciprocal formula: If \(u(t)\) has a derivative and \(u(t) \neq 0\) and $$ P(t)=\frac{1}{u(t)} $$ then $$ P^{\prime}(t)=\frac{-1}{u^{2}(t)} u^{\prime}(t) $$ Prove the formula using logarithmic differentiation. That is, write $$ \ln P(t)=\ln \left(\frac{1}{u(t)}\right)=-\ln u(t) $$ and compute the derivatives of both sides using the logarithm chain rule. We write the formula as $$ \left[\frac{1}{u(t)}\right]^{\prime}=\frac{-1}{u^{2}(t)} u^{\prime}(t) \quad \text { Reciprocal Rule } $$
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