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Chris McMullen - 50 Challenging Calculus Problems (Fully Solved)

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These 50 challenging calculus problems involve applying a variety of calculus skills. The exercises come with a good range of difficulty from milder challenges to very hard problems. On the page following each problem you can find the full solution with explanations.
  • derivatives of polynomials, trig functions, exponentials, and logarithms
  • the chain rule, product rule, and quotient rule
  • second derivatives (and beyond)
  • applications such as related rates, extreme values, and optimization
  • limits, including lHopitals rule
  • antiderivatives of polynomials, trig functions, exponentials, and logarithms
  • definite and indefinite integrals
  • techniques of integration, including substitution, trig sub, and integration by parts
  • multiple integrals
  • non-Cartesian coordinate systems

50 Challenging Calculus Problems (Fully Solved) — read online for free the complete book (whole text) full work

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50 CHALLENGING CALCULUS PROBLEMS Fully Solved Improve Your Math Fluency - photo 1
50 CHALLENGING
CALCULUS PROBLEMS
(Fully Solved)
Improve Your Math Fluency Chris McMullen PhD Copyright 2018 Chris McMullen - photo 2
Improve Your Math Fluency
Chris McMullen, Ph.D.
Copyright 2018 Chris McMullen, Ph.D.
monkeyphysicsblog.wordpress.com
improveyourmathfluency.com
chrismcmullen.com
All rights reserved. However, teachers or parents who purchase one copy of this workbook (or borrow one physical copy from a library) may make and distribute photocopies of selected pages for instructional (non-commercial) purposes for their own students or children only.
Zishka Publishing ISBN 978-1-941691-26-7 Textbooks gt Math gt Calculus - photo 3
Zishka Publishing
ISBN: 978-1-941691-26-7
Textbooks > Math > Calculus
Study Guides > Workbooks> Math
Education > Math > Calculus
Contents
Problem 1
You can find the solution on the following page Solution to Problem 1 This is - photo 4
You can find the solution on the following page.
Solution to Problem 1
This is an application of the chain rule, involving inside and outside functions. After you take a derivative of the outside function, you then take a derivative of the inside function.
According to the chain rule we multiply the two derivatives together To - photo 5
According to the chain rule, we multiply the two derivatives together:
To take the derivative of the cotangent function we will apply the chain rule - photo 6
To take the derivative of the cotangent function, we will apply the chain rule again:
Apply the chain rule again - photo 7
Apply the chain rule again:
Problem 2 Directions Perform the following integral as instructed in - photo 8
Problem 2 Directions Perform the following integral as instructed in each - photo 9
Problem 2 Directions Perform the following integral as instructed in each - photo 10
Problem 2 Directions Perform the following integral as instructed in each - photo 11
Problem 2
Directions : Perform the following integral as instructed in each part below.
A Multiply 2 3x by itself and integrate each term separately B Integrate - photo 12
(A) Multiply 2 3x by itself and integrate each term separately.
(B) Integrate using the method of substitution with u = 2 3x.
(C) Compare your answers for parts (A) and (B). Explain any differences.
You can find the solution on the following page.
Solution to Problem 2
(A) First multiply 2 3x by itself:
(2 3x)
= (2 3x)(2 3x)
= 4 6x 6x + 9x
= 4 12x + 9x
Substitute this into the integral:
As with all indefinite integrals there is an arbitrary constant of - photo 13
As with all indefinite integrals there is an arbitrary constant of - photo 14
As with all indefinite integrals, there is an arbitrary constant of integration, c.
Problem 3 Directions Perform the following eighth derivative - photo 15
Problem 3 Directions Perform the following eighth derivative - photo 16
Problem 3 Directions Perform the following eighth derivative Note that - photo 17
Problem 3 Directions Perform the following eighth derivative Note that - photo 18
Problem 3
Directions : Perform the following eighth derivative.
Note that n read as n factorial means n nn 1n 2321 meaning to - photo 19
Note that n! (read as n factorial) means
n! = n(n 1)(n 2)(3)(2)(1),
meaning to multiply n by all of the integers less than n until you reach the number 1. The eighth derivative is similar to a second derivative, except that it involves taking additional derivatives.
You can find the solution on the following page.
Solution to Problem 3
We need to take eight consecutive derivatives of the given polynomial. It may help to work out the first few derivatives:
Do you have questions about this Following are some answers - photo 20
Do you have questions about this Following are some answers Problem - photo 21
Do you have questions about this Following are some answers Problem 4 - photo 22
Do you have questions about this? Following are some answers:
Problem 4 Directions Perform the following integral You can find the - photo 23
Problem 4 Directions Perform the following integral You can find the - photo 24
Problem 4
Directions : Perform the following integral.
You can find the solution on the following page Solution to Problem 4 - photo 25
You can find the solution on the following page.
Solution to Problem 4
50 Challenging Calculus Problems Fully Solved - photo 26
50 Challenging Calculus Problems Fully Solved - photo 27
Problem 5 Directions Evaluate the following limit - photo 28
Problem 5 Directions Evaluate the following limit You ca - photo 29
Problem 5 Directions Evaluate the following limit You can find the - photo 30
Problem 5 Directions Evaluate the following limit You can find the - photo 31
Problem 5
Directions : Evaluate the following limit.
You can find the solution on the following page Solution to Problem 5 The - photo 32
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