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: 56


$$\frac{{dy}}{{dx}} = - 2{x^3}{\csc ^2}\left( {\frac{1}{2}{x^4}} \right)$$

Work Step by Step

$$\eqalign{ & y = \cot \left( {\frac{1}{2}{x^4}} \right) \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\cot \left( {\frac{1}{2}{x^4}} \right)} \right] \cr & {\text{using the chain rule for }}{D_x}\left( {\cot u} \right) = - {\csc ^2}u \cdot {D_x}\left( u \right).{\text{ for this exercise }}u = \frac{1}{2}{x^4} \cr & {\text{then}} \cr & \frac{{dy}}{{dx}} = - {\csc ^2}\left( {\frac{1}{2}{x^4}} \right)\frac{d}{{dx}}\left[ {\frac{1}{2}{x^4}} \right] \cr & {\text{solve the derivative}} \cr & \frac{{dy}}{{dx}} = - {\csc ^2}\left( {\frac{1}{2}{x^4}} \right)\left( {2{x^3}} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = - 2{x^3}{\csc ^2}\left( {\frac{1}{2}{x^4}} \right) \cr} $$
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