Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 32

Answer

$$f'(a)=6a^2+1$$

Work Step by Step

$$f(t)=2t^3+t$$ That means $$f(a)=2a^3+a$$ According to definition, the derivative of $f(t)$ at number $a$ is $$f'(a)=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$f'(a)=\lim\limits_{h\to0}\frac{[2(a+h)^3+(a+h)]-[2a^3+a]}{h}$$ $$f'(a)=\lim\limits_{h\to0}\frac{[(2a^3+2h^3+6a^2h+6ah^2)+(a+h)]-2a^3-a}{h}$$ $$f'(a)=\lim\limits_{h\to0}\frac{2h^3+6a^2h+6ah^2+h}{h}$$ $$f'(a)=\lim\limits_{h\to0}(2h^2+6ah+6a^2+1)$$ $$f'(a)=2\times0^2+6a\times0+6a^2+1$$ $$f'(a)=6a^2+1$$
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