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: 30

Answer

$$D'(t)=\frac{-3-16t^2}{64t^4}$$

Work Step by Step

$$D(t)=\frac{1+16t^2}{(4t)^3}$$ $$D(t)=\frac{1+16t^2}{64t^3}$$ $$D(t)=\frac{1}{64t^3}+\frac{1}{4t}$$ $$D(t)=\frac{1}{64}t^{-3}+\frac{1}{4}t^{-1}$$ Therefore, the derivative of $D(t)$ is $$D'(t)=\frac{1}{64}\frac{d}{dt}(t^{-3})+\frac{1}{4}\frac{d}{dt}(t^{-1})$$ $$D'(t)=\frac{1}{64}\times(-3)t^{-4}+\frac{1}{4}\times(-1)t^{-2}$$ $$D'(t)=\frac{-3}{64t^4}-\frac{1}{4t^2}$$ $$D'(t)=\frac{-3-16t^2}{64t^4}$$
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