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: 8

Answer

The derivative is $\frac{6t^2+30t+2}{(2t+5)^2}$

Work Step by Step

Using the quotient rule: $g'(t)=(\frac{u(t)}{v(t)})'=\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$ $u(t)=3t^2-1 ;u'(t)=6t$ $v(t)= 2t+5;v'(t)=2$ $g'(t)=\frac{(6t)(2t+5)-(3t^2-1)(2)}{(2t+5)^2}=\frac{12t^2+30t-6t^2+2}{(2t+5)^2}=$ $\frac{6t^2+30t+2}{(2t+5)^2}$
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