Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises: 33

Answer

$f'(x)=\frac{-2x^2+2x-3}{(x^2-3x)^2}$

Work Step by Step

$f(x)=\frac{\frac{2x}{x}-\frac{1}{x}}{x-3}=\frac{2x-1}{x(x-3)}=\frac{2x-1}{x^2-3x}$ Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=2x-1; u'(x)=2$ $v(x)=x^2-3x; v'(x)=2x-3$ $f'(x)=\frac{(2)(x^2-3x)-(2x-3)(2x-1)}{(x^2-3x)^2}$ $=\frac{2x^2-6x-(4x^2-8x+3)}{(x^2-3x)^2}$ $=\frac{-2x^2+2x-3}{(x^2-3x)^2}$
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