University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 37


$ \frac {-2s^2}{\sqrt{1-s^2}}$

Work Step by Step

Given that $y$ = $s\sqrt{{1-s^2}}+cos^{-1}{s}$, on applying the product rule of the derivative, we get: $\frac{dy}{ds}$ = ${\sqrt{1-s^2}\frac{ds}{ds}\times{s\frac{d{\sqrt{{1-s^2}}}}{ds}}\times-\frac{1}{\sqrt{1-s^2}}}$ or $\frac{dy}{ds}$ = $\sqrt{1-s^2} \times { \frac {-2s^2}{2\sqrt{1-s^2}}\times-\frac{1}{\sqrt{1-s^2}}}$ oso $\frac{dy}{ds}$ = $\frac{1-s^2}{\sqrt{1-s^2}} \times { \frac {-s^2}{\sqrt{1-s^2}}\times-\frac{1}{\sqrt{1-s^2}}}$ The final answer is: = $ \frac {-2s^2}{\sqrt{1-s^2}}$
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