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 125: 41

Answer

a. $13$ b. $-7$ c. $\frac{7}{25}$ d. $20$

Work Step by Step

Given $u(0)=5, u'(0)=-3, v(0)=-1, v'(0)=2$, we have: a. $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}=(uv'+vu' )|_{x=0}=5(2)+(-1)(-3)=13$ b. $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}=(\frac{vu'-uv'}{v^2})|_{x=0}=\frac{(-1)(-3)-(5)(2)}{(-1)^2}=-7$ c. $\frac{d}{dx}(\frac{v}{u})=\frac{u\frac{dv}{dx}-v\frac{du}{dx}}{u^2}=(\frac{uv'-vu'}{u^2})|_{x=0}=\frac{(5)(2)-(-1)(-3)}{(5)^2}=\frac{7}{25}$ d. $\frac{d}{dx}(7v-2u)=(7v'-2u')|_{x=0}=7(2)-2(-3)=20$
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