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: 24

Answer

$$G'(t)=\frac{\sqrt 5}{2\sqrt t}-\frac{\sqrt7}{t^2}$$

Work Step by Step

$$G(t)=\sqrt{5t}+\frac{\sqrt7}{t}$$$$G(t)=\sqrt5t^{1/2}+\sqrt7t^{-1}$$ So, $$G'(t)=\sqrt5\frac{d}{dt}(t^{1/2})+\sqrt7\frac{d}{dt}(t^{-1})$$ $$G'(t)=\sqrt5\times\frac{1}{2}\times t^{-1/2}+\sqrt7\times(-1)\times t^{-2}$$ $$G'(t)=\frac{\sqrt5}{2}\times\frac{1}{\sqrt t}-\sqrt 7\times\frac{1}{t^2}$$ $$G'(t)=\frac{\sqrt 5}{2\sqrt t}-\frac{\sqrt7}{t^2}$$
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