## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises: 19

#### Answer

1) Rewrite $\cot x$ into $\frac{\cos x}{\sin x}$ 2) Apply the Quotient Rule for $\frac{d}{dx}\Bigg(\frac{\cos x}{\sin x}\Bigg)$ 3) Use trigonometrical rules to simplify.

#### Work Step by Step

We know that $\cot x=\frac{\cos x}{\sin x}$. So, $$\frac{d}{dx}(\cot x)=\frac{d}{dx}\Bigg(\frac{\cos x}{\sin x}\Bigg)$$ Now apply the Quotient Rule, we have $$\frac{d}{dx}(\cot x)=\frac{(\cos x)'\sin x-\cos x(\sin x)'}{(\sin x)^2}$$ $$\frac{d}{dx}(\cot x)=\frac{(-\sin x)\sin x-\cos x(\cos x)}{\sin^2 x}$$ $$\frac{d}{dx}(\cot x)=\frac{-\sin^2 x-\cos^2 x}{\sin^2 x}$$ $$\frac{d}{dx}(\cot x)=\frac{-(\sin^2 x+\cos^2 x)}{\sin^2 x}$$ $$\frac{d}{dx}(\cot x)=\frac{-1}{\sin^2 x}$$ $$\frac{d}{dx}(\cot x)=-\csc^2 x$$ The statement has been proved.

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