Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 46

Answer

- First derivative: $$G'(r)=\frac{1}{2\sqrt r}+\frac{1}{3\sqrt[3]{r^2}}$$ - Second derivative: $$G''(r)=-\frac{1}{4\sqrt{r^3}}-\frac{2}{9\sqrt[3]{r^5}}$$

Work Step by Step

$$G(r)=\sqrt r+\sqrt[3]r$$$$G(r)=r^{1/2}+r^{1/3}$$ - First derivative: $$G'(r)=\frac{d}{dr}(r^{1/2})+\frac{d}{dr}(r^{1/3})$$ $$G'(r)=\frac{1}{2}r^{-1/2}+\frac{1}{3}r^{-2/3}$$ $$G'(r)=\frac{1}{2r^{1/2}}+\frac{1}{3r^{2/3}}$$ $$G'(r)=\frac{1}{2\sqrt r}+\frac{1}{3\sqrt[3]{r^2}}$$ - Second derivative: $$G''(r)=\frac{1}{2}\frac{d}{dr}(r^{-1/2})+\frac{1}{3}\frac{d}{dr}(r^{-2/3})$$ $$G''(r)=\frac{1}{2}\times(-\frac{1}{2})r^{-3/2}+\frac{1}{3}\times(-\frac{2}{3})r^{-5/3}$$ $$G''(r)=-\frac{1}{4}\times\frac{1}{\sqrt{r^3}}-\frac{2}{9}\times\frac{1}{\sqrt[3]{r^5}}$$ $$G''(r)=-\frac{1}{4\sqrt{r^3}}-\frac{2}{9\sqrt[3]{r^5}}$$
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