Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.6 Inverse Trigonometric Functions - 6.6 Exercises - Page 481: 29

Answer

$$ F^{\prime}(x) = \frac{d}{dx} \left[ x \sec ^{-1}\left(x^{3}\right) \right] =\frac{3}{\sqrt{x^{6}-1}}+\sec ^{-1}\left(x^{3}\right) $$

Work Step by Step

$$ F(x)=x \sec ^{-1}\left(x^{3}\right) $$ Differentiating both sides of this equation, using multiply Rule we have $$ \begin{aligned} \frac{d}{dx} \left[ F(x) \right] &= \frac{d}{dx} \left[ x \sec ^{-1}\left(x^{3}\right) \right] \\ F^{\prime}(x) &=x \cdot \frac{1}{x^{3} \sqrt{\left(x^{3}\right)^{2}-1}} \frac{d}{d x}\left(x^{3}\right)+\sec ^{-1}\left(x^{3}\right) \cdot 1 \\ &=\frac{x\left(3 x^{2}\right)}{x^{3} \sqrt{x^{6}-1}}+\sec ^{-1}\left(x^{3}\right) \\ &=\frac{3}{\sqrt{x^{6}-1}}+\sec ^{-1}\left(x^{3}\right) \end{aligned} $$
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