Calculus with Applications (10th Edition)

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

Chapter 13 - The Trigonometric Functions - 13.2 Derivatives of Trigonometric Functions - 13.2 Exercises - Page 689: 22

Answer

$$\frac{{dy}}{{dx}} = - \frac{{15\sin x}}{{{{\left( {5 - \cos x} \right)}^2}}}$$

Work Step by Step

$$\eqalign{ & y = \frac{{3\cos x}}{{5 - \cos x}} \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = {D_x}\left[ {\frac{{3\cos x}}{{5 - \cos x}}} \right] \cr & {\text{use the quotient rule for derivatives}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {5 - \cos x} \right) \cdot {D_x}\left( {3\cos x} \right) - 3\cos x \cdot {D_x}\left( {5 - \cos x} \right)}}{{{{\left( {5 - \cos x} \right)}^2}}} \cr & {\text{solve derivatives }}{D_x}\left( {\cos x} \right) = - \sin x \cr & \frac{{dy}}{{dx}} = \frac{{\left( {5 - \cos x} \right)\left( { - 3\sin x} \right) - 3\cos x\left( {\sin x} \right)}}{{{{\left( {5 - \cos x} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{ - 15\sin x + 3\cos x\sin x - 3\cos x\sin x}}{{{{\left( {5 - \cos x} \right)}^2}}} \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = - \frac{{15\sin x}}{{{{\left( {5 - \cos x} \right)}^2}}} \cr} $$
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