University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises - Page 137: 76


The detailed proof is below.

Work Step by Step

$$\frac{d}{dx}(x^{-m})=\frac{d}{dx}\Big(\frac{1}{x^m}\Big)$$ Apply the Derivative Quotient Rule here: $$\frac{d}{dx}(x^{-m})=\frac{(1)'x^m-1(x^m)'}{(x^m)^2}$$ $$\frac{d}{dx}(x^{-m})=\frac{0\times x^m-mx^{m-1}}{x^{2m}}$$ $$\frac{d}{dx}(x^{-m})=\frac{-mx^{m-1}}{x^{2m}}$$ $$\frac{d}{dx}(x^{-m})=-mx^{m-1-2m}=-mx^{-m-1}$$ The Power Rule for negative integers has been proved.
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