## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.7 Derivatives And Integrals Involving Inverse Trigonometric Functions - Exercises Set 6.7 - Page 469: 19

#### Answer

$$\frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{1 + {x^6}}}$$

#### Work Step by Step

\eqalign{ & y = {\tan ^{ - 1}}\left( {{x^3}} \right) \cr & {\text{differentiate using the formula }} \cr & \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr & \cr & \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {{x^3}} \right)}^2}}}\frac{d}{{dx}}\left( {{x^3}} \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {{x^3}} \right)}^2}}}\left( {3{x^2}} \right) \cr & {\text{simplifying}}{\text{, }} \cr & \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{1 + {x^{\left( 3 \right)\left( 2 \right)}}}} \cr & \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{1 + {x^6}}} \cr}

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