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: 19

Answer

$r'=\frac{cos((2θ)^{1/2})}{(2θ)^{1/2}}$

Work Step by Step

Rewrite the equation: $r=sin((2θ)^{1/2})$ Take the derivative of the equation using Trigonometric derivative and Chain Rule: $r'=cos((2θ)^{1/2})\times\frac{1}{2}(2θ)^{-1/2}\times2$ $=\frac{cos((2θ)^{1/2})}{(2θ)^{1/2}}$
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