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: 6

Answer

The derivative is $\frac{(sin(x)+2xcos(x))(\sqrt x)}{2x}$.

Work Step by Step

Product Rule $g'(x)=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=\sqrt x ;u’(x)=\frac{1}{2\sqrt x} $ $v(x)=\sin(x) ;v’(x)=\cos(x)$ $g'(x)=(\frac{1}{2\sqrt x})(\sin(x))+(\sqrt x)(cos(x))=$ $(\frac{1}{2\sqrt x})(\sin(x))+(\frac{2x}{2\sqrt x})(cos(x))=$ $(\frac{1}{2\sqrt x})(sin(x)+2xcos(x))=$ $\frac{sin(x)+2xcos(x)}{2\sqrt x}=$ $\frac{(sin(x)+2xcos(x))(\sqrt x)}{2x}$
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