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: 33



Work Step by Step

$$f(t)=\frac{2t+1}{t+3}$$ That means $$f(a)=\frac{2a+1}{a+3}$$ The derivative of $f(t)$ at number $a$ can be found as follows: $$f'(a)=\lim\limits_{t\to a}\frac{f(t)-f(a)}{t-a}$$ $$f'(a)=\lim\limits_{t\to a}\frac{\frac{2t+1}{t+3}-\frac{2a+1}{a+3}}{t-a}$$ $$f'(a)=\lim\limits_{t\to a}\frac{\frac{(2t+1)(a+3)-(2a+1)(t+3)}{(t+3)(a+3)}}{t-a}$$ $$f'(a)=\lim\limits_{t\to a}\frac{(2at+6t+a+3)-(2at+6a+t+3)}{(t+3)(a+3)(t-a)}$$ $$f'(a)=\lim\limits_{t\to a}\frac{5t-5a}{(t+3)(a+3)(t-a)}$$ $$f'(a)=\lim\limits_{t\to a}\frac{5}{(t+3)(a+3)}$$ $$f'(a)=\frac{5}{(a+3)(a+3)}$$ $$f'(a)=\frac{5}{(a+3)^2}$$
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