Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.2 The Derivative Function - Exercises Set 2.2: 17

Answer

$\frac{dy}{dx} = 2x-1 $

Work Step by Step

$y = x^2 - x $ $\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {f(x+∆x) -f(x)}{∆x} $ $\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {[(x+∆x)^2 - (x+∆x)]-[x^2-x]}{∆x} $ $\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {x^2+2x∆x+(∆x)^2-x-∆x-x^2+x}{∆x} $ $\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {2x∆x+(∆x)^2 -∆x}{∆x} $ $\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {∆x(2x+∆x-1)}{∆x} $ $\frac{dy}{dx} = \lim\limits_{∆x \to 0} 2x+∆x-1 $ $\frac{dy}{dx} = 2x+0-1 $ $\frac{dy}{dx} = 2x-1 $
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