Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises: 49

Answer

$y'=\cos{x}\cot^2{x}.$

Work Step by Step

$y=f(x)+g(x)\rightarrow f(x)=-\csc{x}$; $g(x)=-\sin{x}$ By Theorem 2.9: $f'(x)$ $=(-1)(-\csc{x}\cot{x})$ $=\csc{x}\cot{x}$ $g'(x)=\dfrac{d}{dx}(-\sin{x})=-\cos{x}$ $y'=f'(x)+g'(x)=\csc{x}\cot{x}-\cos{x}=\cos{x}\cot^2{x}.$
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