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 - Page 126: 64

Answer

See explanations.

Work Step by Step

Step 1. Recall the Derivative Quotient Rule: $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$ Step 2. Let $u=1$, we have: $\frac{d}{dx}(\frac{1}{v})=\frac{0-\frac{dv}{dx}}{v^2}=-\frac{1}{v^2}(\frac{dv}{dx})$ Step 3. Let $v=x^m$, we have $\frac{d}{dx}(x^{-m})=\frac{d}{dx}(\frac{1}{x^m})=-\frac{1}{(x^m)^2}(\frac{dx^m}{dx})=-\frac{mx^{m-1}}{(x^m)^2}=-mx^{m-1-2m}=-mx^{-m-1}$ which gives the Power Rule for negative integers.
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