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: 42



Work Step by Step

Given that $y$ = $\ln{(x^2+4)}+x\tan^{-1}{\frac{x}{2}}$, on applying the product rule and chain rule of the derivative, we get: $\frac{dy}{dx}$ = ${\frac{1}{x^2+4}\times\frac{d(x^2+4)}{dx}}-({tan^{-1}{\frac{x}{2}+x{\frac{d\tan^{-1}{\frac{x}{2}}}{dx}}}})$ or $\frac{dy}{dx}$ = ${\frac{2x}{x^2+4}-\tan^{-1}{\frac{x}{2}}-x{\frac{1}{1+{\frac{x^2}{4}}}{\frac{1}{2}}}}$ $\frac{dy}{dx}$ = ${\frac{2x}{x^2+4}-\tan^{-1}{\frac{x}{2}}-{\frac{2x}{x^2+4}}}$ The final answer is:$\frac{dy}{dx} = -\tan^{-1}{\frac{x}{2}}$
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