Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.4 Exercises: 51

Answer

$f'(x)=-\csc{x}(\csc^2{x}+\cot^2{x}).$

Work Step by Step

$f(x)=\dfrac{\cot{x}}{\sin{x}}=\cot{x}\csc{x}.$ Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=\cot{x} ;u’(x)=-\csc^2{x} $. $v(x)=\csc{x} ;v’(x)=-\csc{x}\cot{x} $. $f'(x)=(-\csc^2{x})(\csc{x})+(-\csc{x}\cot{x})(\cot{x})$ $=-\csc{x}(\csc^2{x}+\cot^2{x})$.
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.