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: 33


$$\frac{d}{{dx}}\left[ {\cot x} \right] = - {\csc ^2}x$$

Work Step by Step

$$\eqalign{ & \cot x = \frac{{\cos x}}{{\sin x}} \cr & {\text{differentiate with respect to }}x \cr & \frac{d}{{dx}}\left[ {\cot x} \right] = \frac{d}{{dx}}\left[ {\frac{{\cos x}}{{\sin x}}} \right] \cr & {\text{by using the quotient rule}} \cr & \frac{d}{{dx}}\left[ {\cot x} \right] = \frac{{\left( {\sin x} \right)\frac{d}{{dx}}\left[ {\cos x} \right] - \cos x\frac{d}{{dx}}\left[ {\sin x} \right]}}{{{{\left( {\sin x} \right)}^2}}} \cr & {\text{then}} \cr & \frac{d}{{dx}}\left[ {\cot x} \right] = \frac{{\left( {\sin x} \right)\left( { - \sin x} \right) - \cos x\left( {\cos x} \right)}}{{{{\left( {\sin x} \right)}^2}}} \cr & {\text{simplifying}} \cr & \frac{d}{{dx}}\left[ {\cot x} \right] = \frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\left( {\sin x} \right)}^2}}} \cr & \frac{d}{{dx}}\left[ {\cot x} \right] = - \frac{1}{{{{\left( {\sin x} \right)}^2}}} \cr & {\text{use }}\csc x = \frac{1}{{\sin x}} \cr & \frac{d}{{dx}}\left[ {\cot x} \right] = - {\csc ^2}x \cr} $$
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