Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 21

Answer

$u'=-\dfrac{2}{t^3}+\dfrac{1}{\sqrt{t^3}}+1$

Work Step by Step

Simplify the expression $u=(t^{-1}-t^{-1/2})^2$ Apply newton's binomial formula: $(a-b)^2=a^2-2ab+b^2$ $u=(t^{-1})^2-2(t^{-1})(t^{-1/2})+(t^{-1/2})^2$ $u=t^{-2}-2t^{-3/2}+t$ Apply power rule to derivate $u'=(-2)t^{-2-1}-2(\dfrac{1}{2})t^{-1/2-1}+(1)t^{1-1}$ $u'=-\dfrac{2}{t^3}+\dfrac{1}{\sqrt{t^3}}+1$
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