/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Suppose \(u\) and \(v\) are func... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 53

Suppose \(u\) and \(v\) are functions of \(x\) that are differentiable at \(x=0\) and that $$u(0)=5, \quad u^{\prime}(0)=-3, \quad v(0)=-1, \quad v^{\prime}(0)=2$$ Find the values of the following derivatives at \(x=0\) $$\text { a. } \frac{d}{d x}(u v)$$ $$\text { b. } \frac{d}{d x}\left(\frac{u}{v}\right)$$ $$\text { c. } \frac{d}{d x}\left(\frac{v}{u}\right)$$ $$\text { d. } \frac{d}{d x}(7 v-2 u)$$

Short Answer

Expert verified
a. 13, b. -7, c. \(\frac{7}{25}\), d. 20

Step by step solution

01

Differentiate Product (Part a)

We need to find \( \frac{d}{dx}(uv) \) using the product rule, which states that if \( u \) and \( v \) are differentiable functions, then \( \frac{d}{dx}(uv) = u'v + uv' \). At \( x = 0 \), substitute the given values: \[ u'(0) = -3, \ u(0) = 5, \ v(0) = -1, \ v'(0) = 2 \] So, \[ \frac{d}{dx}(uv)\bigg|_{x=0} = (-3)(-1) + (5)(2) = 3 + 10 = 13 \]
02

Differentiate Quotient (Part b)

To find \( \frac{d}{dx}\left(\frac{u}{v}\right) \), use the quotient rule: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] Substitute the given values at \( x = 0 \): \[ \frac{d}{dx}\left(\frac{u}{v}\right)\bigg|_{x=0} = \frac{(-3)(-1) - (5)(2)}{(-1)^2} = \frac{3 - 10}{1} = -7 \]
03

Differentiate Reciprocal (Part c)

To find \( \frac{d}{dx}\left(\frac{v}{u}\right) \), also use the quotient rule: \[ \frac{d}{dx}\left(\frac{v}{u}\right) = \frac{v'u - uv'}{u^2} \] Substitute the given values: \[ \frac{d}{dx}\left(\frac{v}{u}\right)\bigg|_{x=0} = \frac{(2)(5) - (-1)(-3)}{5^2} = \frac{10 - 3}{25} = \frac{7}{25} \]
04

Differentiate Linear Combination (Part d)

We need to find \( \frac{d}{dx}(7v - 2u) \). The derivative of a linear combination is simply the linear combination of the derivatives: \[ \frac{d}{dx}(7v - 2u) = 7v' - 2u' \] Using the given values: \[ \frac{d}{dx}(7v - 2u)\bigg|_{x=0} = 7(2) - 2(-3) = 14 + 6 = 20 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Product Rule
When you're working with the product of two functions, such as \(u(x)v(x)\), the product rule comes in handy for differentiation. The rule states: \[\frac{d}{dx}(uv) = u'v + uv'\] This means you'll take the derivative of the first function, multiply it by the second function, and then add it to the first function multiplied by the derivative of the second function. Here's a breakdown of how it works:
  • Find the derivative of function \(u\), which is \(u'\).
  • Multiply \(u'\) by the function \(v\).
  • Find the derivative of function \(v\), which is \(v'\).
  • Multiply \(v'\) by the function \(u\).
  • Add these two results together to get the final derivative.
So, the combined effort of differentiating and adding these components gives you the derivative of the product of two functions.
Quotient Rule
The quotient rule is what you use when differentiating a function that's the ratio of two differentiable functions, such as \(\frac{u}{v}\). The quotient rule is given by:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\]This rule requires you to:
  • Differentiate the numerator to get \(u'\).
  • Differentiate the denominator to get \(v'\).
  • Multiply \(u'\) by \(v\) and \(u\) by \(v'\).
  • Subtract the second product from the first.
  • Divide everything by the square of \(v\).
This approach ensures that you are systematically finding the rate of change of a quotient of two functions.
Reciprocal Function
The concept of a reciprocal function ties closely to the quotient rule. If you have \(\frac{v}{u}\), and you need to differentiate it, you will again employ the quotient rule but apply it to the inverse form: \[\frac{d}{dx}\left(\frac{v}{u}\right) = \frac{v'u - uv'}{u^2}\]Key things to remember:
  • You are essentially considering the function as a ratio of two differentiable functions, just reversed.
  • Find derivatives just like with the standard quotient rule.
  • Pay careful attention to sign changes that can occur in subtraction.
  • Divide by the square of the denominator function.
This makes differentiating reciprocal functions systematic and consistent, relying on foundational calculus rules.
Linear Combination
Linear combinations involve sums of functions multiplied by constants. When differentiating a linear combination like \(7v - 2u\), it is quite straightforward:The rule is: \[\frac{d}{dx}(7v - 2u) = 7v' - 2u'\] Here's how it works:
  • The derivative of a constant times a function is the constant times the derivative of the function.
  • Differentiate each function in the combination independently.
  • Multiply the resulting derivatives by their respective constants.
  • Combine the results using addition or subtraction exactly as in the original function.
This shows how differentiation respects the linearity of functions, making these calculations simpler to perform.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=t(t+1)(t+2)$$

a. Let \(Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2}\) be a quadratic approximation to \(f(x)\) at \(x=a\) with the properties: i) \(Q(a)=f(a)\) ii) \(Q^{\prime}(a)=f^{\prime}(a)\) iii) \(Q^{\prime \prime}(a)=f^{\prime \prime}(a)\) Determine the coefficients \(b_{0}, b_{1},\) and \(b_{2}\) b. Find the quadratic approximation to \(f(x)=1 /(1-x)\) at \(x=0\) c. Graph \(f(x)=1 /(1-x)\) and its quadratic approximation at \(x=0 .\) Then zoom in on the two graphs at the point (0,1) Comment on what you see. d. Find the quadratic approximation to \(g(x)=1 / x\) at \(x=1\) Graph \(g\) and its quadratic approximation together. Comment on what you see. e. Find the quadratic approximation to \(h(x)=\sqrt{1+x}\) at \(x=0 .\) Graph \(h\) and its quadratic approximation together. Comment on what you see. f. What are the linearizations of \(f, g,\) and \(h\) at the respective points in parts (b), (d), and (e)?

Find the derivative of \(y\) with respect to the given independent variable. $$y=x^{\pi}$$

Use the identity $$\csc ^{-1} u=\frac{\pi}{2}-\sec ^{-1} u$$ to derive the formula for the derivative of \(\csc ^{-1} u\) in Table 3.1 from the formula for the derivative of \(\sec ^{-1} u\)

Graph the function \(y=2 \cos 2 x\) for \(-2 \leq x \leq 3.5 .\) Then, on the same screen, graph $$y=\frac{\sin 2(x+h)-\sin 2 x}{h}$$ for \(h=1.0,0.5,\) and \(0.2 .\) Experiment with other values of \(h\) including negative values. What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.