Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.5 Derivatives of Trigonometric Functions - 3.5 Exercises - Page 169: 31

Answer

\[ = - \csc x\cot x\]

Work Step by Step

\[\begin{gathered} \hfill \\ \frac{d}{{dx}}\,\left( {\csc x} \right) = - \csc x\cot x \hfill \\ \hfill \\ use\,\,trigonometric\,\,identity\,\,\csc x = \frac{1}{{\sin x}} \hfill \\ then \hfill \\ \hfill \\ \frac{d}{{dx}}\,\left( {\frac{1}{{\sin x}}} \right) = \frac{{0 - \,\left( {\cos x} \right)}}{{{{\sin }^2}x}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \,\left( { - \frac{{\cos x}}{{\sin x}}} \right)\,\left( {\frac{1}{{\sin x}}} \right) \hfill \\ \hfill \\ = - \csc x\cot x \hfill \\ \hfill \\ \end{gathered} \]
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