## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 38

#### Answer

${ \frac {(s|s|-1)}{|s|\sqrt{s^2-1}}}$ or ${ \frac {(s^2-1)}{s\sqrt{s^2-1}}}$

#### Work Step by Step

Given that $y$ = $\sqrt{{s^2-1}}-sec^{-1}{s}$, on applying the product rule of the derivative, we get: $\frac{dy}{ds}$ = ${\frac{1}{2}\frac{}{\sqrt{s^2-1}}\frac{d({s^2-1})}{ds}-\frac{1}{|s|\sqrt{s^2-1}}}$ or $\frac{dy}{ds}$ = ${ \frac {2s}{2\sqrt{s^2-1}}-{\frac{1}{|s|\sqrt{(s^2-1)}}}}$ On simplifying: $\frac{dy}{ds}$ = ${ \frac {(s|s|-1)}{|s|\sqrt{s^2-1}}}$ or ${ \frac {(s^2-1)}{s\sqrt{s^2-1}}}$ The final answer is: = ${ \frac {(s|s|-1)}{|s|\sqrt{s^2-1}}}$ or ${ \frac {(s^2-1)}{s\sqrt{s^2-1}}}$

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