Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Practice Exercises - Page 177: 24

Answer

$y'=-3x^{1/2}(x+1)^2csc(x+1)^3cot(x+1)^3+\frac{1}{2}x^{-1/2}csc(x+1)^3$

Work Step by Step

Rewrite the equation: $y=x^{1/2}csc(x+1)^3$ Take the derivative of the equation using Power Rule, Product Rule, Trigonometric derivatives, and Chain Rule: $y'=x^{1/2}\times-csc(x+1)^3cot(x+1)^3\times3(x+1)^2\times1+\frac{1}{2}x^{-1/2}csc(x+1)^3$ $=-3x^{1/2}(x+1)^2csc(x+1)^3cot(x+1)^3+\frac{1}{2}x^{-1/2}csc(x+1)^3$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.