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 688: 14

Answer

$$\frac{{dy}}{{dx}} = - 8{e^{2x}}\sin \left( {4{e^{2x}}} \right)$$

Work Step by Step

$$\eqalign{ & y = \cos \left( {4{e^{2x}}} \right) \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = {D_x}\left[ {\cos \left( {4{e^{2x}}} \right)} \right] \cr & {\text{use the chain rule }}{D_x}\left( {\cos u} \right) = - \sin u \cdot {D_x}\left( u \right).{\text{ consider }}u = 4{e^{2x}} \cr & \frac{{dy}}{{dx}} = - \sin \left( {4{e^{2x}}} \right) \cdot {D_x}\left( {4{e^{2x}}} \right) \cr & \frac{{dy}}{{dx}} = - 4\sin \left( {4{e^{2x}}} \right) \cdot {D_x}\left( {{e^{2x}}} \right) \cr & {\text{use the chain rule }}{D_x}\left( {{e^u}} \right) = {e^u} \cdot {D_x}\left( u \right) \cr & \frac{{dy}}{{dx}} = - 4\sin \left( {4{e^{2x}}} \right)\left( {{e^{2x}}} \right) \cdot {D_x}\left( {2x} \right) \cr & {\text{solve the derivative and simplify}} \cr & \frac{{dy}}{{dx}} = - 4\sin \left( {4{e^{2x}}} \right)\left( {{e^{2x}}} \right)\left( 2 \right) \cr & \frac{{dy}}{{dx}} = - 8{e^{2x}}\sin \left( {4{e^{2x}}} \right) \cr} $$
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