Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises: 33

Answer

$f'(a) = \dfrac{5}{(a+3)^2}$

Work Step by Step

Given $f(t) = \dfrac{2t+1}{t+3}$ $f'(a) = \lim\limits_{t \to a} \dfrac{f(t) - f(a)}{t - a} = \lim\limits_{t \to a} \dfrac{\frac{2t+1}{t+3} - \frac{2a+1}{a+3}}{t - a} = \lim\limits_{t \to a} \dfrac{\frac{(2t+1)(a+3) - (2a+1)(t+3)}{(t+3)(a+3)}}{t - a} = \lim\limits_{t \to a} \dfrac{(2t+1)(a+3) - (2a+1)(t+3)}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{2at+a+6t+3 - (2at+t+6a+3)}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{2at+a+6t+3 - 2at-t-6a-3}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{a+6t-t-6a}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{5t-5a}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{5(t-a)}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{5}{(t+3)(a+3)} = \dfrac{5}{(a+3)(a+3)} = \dfrac{5}{(a+3)^2}\longrightarrow f'(a) = \dfrac{5}{(a+3)^2}$
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