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: 3

Answer

The derivative is $\frac{(1-5t^2)(\sqrt t)}{2t}$

Work Step by Step

Product Rule $h'(t)=((u(t)(v(t))’=u'(t)v(t)+u(t)v’(t))$ $u(t)=\sqrt t ;u’(t)= \frac{1}{2\sqrt t}$ $v(t)=(1-t^2) ;v’(t)=-2t $ $h'(t)= (\frac{1}{2\sqrt t})(1-t^2)+(\sqrt t)((-2t)=$ $h'(t)= (\frac{1}{2\sqrt t})(1-t^2)+(\frac{2t}{2\sqrt t})((-2t)=$ $(\frac{1}{2\sqrt t})((1-t^2)+-4t^2)=$ $\frac{1-5t^2}{2\sqrt t}=\frac{(\sqrt t)(1-5t^2)}{2t}$
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