Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises: 4

Answer

The derivative is $\frac{(5s^2+8)(\sqrt s)}{2s}$

Work Step by Step

Product Rule $g'(s)=((u(s)(v(s))’=u’(s)v(s)+u(s)v’(s))$ $u(s)=\sqrt s ;u’(s)=\frac{1}{2\sqrt s} $ $v(s)=(s^2+8) ;v’(s)=2s $ $g'(s)= (\frac{1}{2\sqrt s})(s^2+8)+(\sqrt s)(2s)=$ $g'(s)= (\frac{1}{2\sqrt s})(s^2+8)+(\frac{2s}{2\sqrt s)}(2s)=$ $(\frac{1}{2\sqrt s})((s^2+8)+4s^2)=$ $\frac{5s^2+8}{2\sqrt s}=\frac{(5s^2+8)(\sqrt s)}{2s}$
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