Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 35

Answer

See the proof below.

Work Step by Step

\begin{align*} \frac{d }{dx} \cot x&=\frac{d }{dx} \frac{\cos x}{\sin x}\\ &=\frac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2 x}\\ &=\frac{-\sin^2 x-\cos^2 x}{\sin^2 x}\\ &=\frac{-1}{\sin^2 x}\\ &=-\csc^2 x. \end{align*}
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