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 178: 36

Answer

$r'=2(\frac{1+sinθ}{1-cosθ})\times\frac{cosθ-1-sinθ}{(1-cosθ)^2}$

Work Step by Step

Take the derivative of the equation using Power Rule, Chain Rule, and Quotient Rule: $r'=2(\frac{1+sinθ}{1-cosθ})\times\frac{(1-cosθ)(0+cosθ)-(1+sinθ)(0+sinθ)}{(1-cosθ)^2}$ $=2(\frac{1+sinθ}{1-cosθ})\times\frac{cosθ-cos^2θ-sinθ-sin^2θ}{(1-cosθ)^2}$ Simplify using the Trigonometric Identity: $cos^2θ+sin^2θ=1$ $=2(\frac{1+sinθ}{1-cosθ})\times\frac{cosθ-1-sinθ}{(1-cosθ)^2}$
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