Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises: 32

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 $h'(x)=1$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.