# Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 421: 42

$$\frac{{dy}}{{dx}} = - {\tan ^{ - 1}}\left( {\frac{x}{2}} \right)$$

#### Work Step by Step

\eqalign{ & y = \ln \left( {{x^2} + 4} \right) - x{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) \cr & {\text{find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{{d\left( {\ln \left( {{x^2} + 4} \right)} \right)}}{{dx}} + \frac{{d\left( {x{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right)}}{{dx}} \cr & {\text{product rule}} \cr & \frac{{dy}}{{dx}} = \frac{{d\left( {\ln \left( {{x^2} + 4} \right)} \right)}}{{dx}} - x\frac{{d\left( {{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right)}}{{dx}} - {\tan ^{ - 1}}\left( {\frac{x}{2}} \right)\frac{{d\left( x \right)}}{{dx}} \cr & {\text{ use the formulas }} \cr & \frac{{d\left( {{{\tan }^{ - 1}}u} \right)}}{{dx}} = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}},.\,\,\left( {{\text{see table 7}}{\text{.3}}} \right){\text{ and the chain rule}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{{x^2} + 4}}\frac{{d\left( {{x^2} + 4} \right)}}{{dx}} - x\left( {\frac{1}{{1 + {{\left( {x/2} \right)}^2}}}} \right)\frac{{d\left( {x/2} \right)}}{{dx}} - {\tan ^{ - 1}}\left( {\frac{x}{2}} \right)\left( 1 \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{{{x^2} + 4}}\left( {2x} \right) - x\left( {\frac{4}{{4 + {x^2}}}} \right)\left( {\frac{1}{2}} \right) - {\tan ^{ - 1}}\left( {\frac{x}{2}} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = \frac{{2x}}{{{x^2} + 4}} - \frac{{2x}}{{4 + {x^2}}} + {\tan ^{ - 1}}\left( {\frac{x}{2}} \right) \cr & \frac{{dy}}{{dx}} = - {\tan ^{ - 1}}\left( {\frac{x}{2}} \right) \cr}

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