## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 141: 41

#### Answer

No, see explanations.

#### Work Step by Step

Step 1. Given $y=x-cot(x)$, we have $y'=1+csc^2x$ Step 2. For a horizontal tangent, let $y'=0$; we have $csc^2(x)=-1$. Step 3. As there is no solution to the equation above, we can not find a horizontal tangent.

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