Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.3 Differentiation Rules - Exercises 3.3: 19

Answer

The Derivative is: $y'=\frac{x^2+\space x\space +\space 4}{(x\space +\space 0.5)^2}$

Work Step by Step

$y=\frac{x^2\space -\space 4}{x\space +\space 0.5}$ Using the Quocient Rule: $y'=\frac{f'(x)\cdot g(x)\space - \space f(x)\cdot g'(x)}{g^2(x)}$ $y'=\frac{((2)x^{2-1} -\space 0)(x\space +\space 0.5)\space -\space (x^2-\space 4)((1)x^{1-1}+\space 0)}{(x\space +\space 0.5)^2}$ $y'=\frac{2x(x\space +\space 0.5)\space -\space (1)(x^2-\space 4)}{(x\space +\space 0.5)^2}$ $y'=\frac{2x^2- x^2+\space x\space+4}{(x\space +\space 0.5)^2}$ $y'=\frac{x^2+\space x\space +\space 4}{(x\space +\space 0.5)^2}$
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