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: 37

Answer

\[\frac{{dy}}{{dx}} = \frac{{ - {{\csc }^2}x - {{\csc }^3}x + {{\cot }^2}x\csc x}}{{\,{{\left( {1 + {{\csc }^2}x} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{\cot x}}{{1 + \csc x}} \hfill \\ \hfill \\ {\text{using}}\,\,the\,\,quotient\,\,rule. \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{\left( {1 + \csc x} \right)\left[ {\cot x} \right]' - \left( {\cot x} \right)\left[ {1 + \csc x} \right]'}}{{{{\left( {1 + \csc x} \right)}^2}}} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{\,\left( {1 + \csc x} \right)\,\left( { - {{\csc }^2}x} \right) - \cot x\,\left( { - \csc x\cot x} \right)}}{{\,{{\left( {1 + \csc x} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{ - {{\csc }^2}x - {{\csc }^3}x + {{\cot }^2}x\csc x}}{{\,{{\left( {1 + {{\csc }^2}x} \right)}^2}}} \hfill \\ \end{gathered} \]
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