Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 13 - The Trigonometric Functions - Chapter Review - Review Exercises - Page 701: 65


$$\frac{{dy}}{{dx}} = \cot x$$

Work Step by Step

$$\eqalign{ & y = \ln \left| {5\sin x} \right| \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = {D_x}\left( {\ln \left| {5\sin x} \right|} \right) \cr & {\text{use the chain rule }}{D_x}\left( {\ln u} \right) = \frac{1}{u} \cdot {D_x}\left( u \right).{\text{ consider }}u = 5\sin x \cr & \frac{{dy}}{{dx}} = \left( {\frac{1}{{5\sin x}}} \right){D_x}\left( {5\sin x} \right) \cr & {\text{solve the derivative }} \cr & \frac{{dy}}{{dx}} = \left( {\frac{1}{{5\sin x}}} \right)\left( {5\cos x} \right) \cr & {\text{then}} \cr & \frac{{dy}}{{dx}} = \frac{{\cos x}}{{\sin x}} \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = \cot x \cr} $$
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