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: 26

Answer

$$\frac{{dy}}{{dx}} = 24{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^7}\left( {\cos 3x - {x^2}{{\csc }^2}{x^3}} \right)$$

Work Step by Step

$$\eqalign{ & y = {\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^8} \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = {D_x}{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^8} \cr & {\text{use the chain rule}} \cr & \frac{{dy}}{{dx}} = 8{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^7}{D_x}\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right] \cr & {\text{use the differentiation rule for trigonometric functions}}{\text{, then}} \cr & \frac{{dy}}{{dx}} = 8{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^7}\left( {3\cos 3x - {{\csc }^2}{x^3}\left( {3{x^2}} \right)} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = 24{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^7}\left( {\cos 3x - {x^2}{{\csc }^2}{x^3}} \right) \cr} $$
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