Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.8 - The Derivative as a Function - 2.8 Exercises: 23



Work Step by Step

$f(t)=2.5t^{2}+6t$ Derivative using the definition: $f'(t)=\lim\limits_{h \to 0}\dfrac{f(t+h)-f(t)}{h}$ To find $f(t+h)$, wherever you see $t$ in the function, plug in $t+h$ $f(t+h)=2.5(t+h)^{2}+6(t+h)=2.5(t^{2}+2th+h^{2})+6t+6h=...$ $...=2.5t^{2}+5th+2.5h^{2}+6t+6h$ Let's plug in the components of the formula: $f'(t)=\lim\limits_{h \to 0}\dfrac{f(t+h)-f(x)}{h}=\lim\limits_{h \to 0}\dfrac{2.5t^{2}+5th+2.5h^{2}+6t+6h-2.5t^{2}-6t}{h}=\lim\limits_{h \to 0}\dfrac{5th+2.5h^{2}+6h}{h}$ Take out common factor $h$ $\lim\limits_{h \to 0}\dfrac{h(5t+2.5h+6)}{h}=\lim\limits_{h \to 0}5t+2.5h+6$ Apply direct substitution: $5t+2.5(0)+6=5t+6$ $f'(t)=5t+6$
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