Answer
$h'(x)=1$
Work Step by Step
Simplify the expression:
$h(x)=\dfrac{x^3-6x^2+8x}{x^2-2x}=\dfrac{x(x^2-6x+8)}{x(x-2)}$
$h(x)=\dfrac{x^2-6x+8}{x-2}=\dfrac{(x-4)(x-2)}{x-2}$
$h(x)=x+4$
Apply power rule
$h'(x)=1$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises - Page 151: 32
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