Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 31


$\frac{t}{\sqrt (t^{2}+9)}$

Work Step by Step

Given y= $(t^{2}+9)^{1/2}$ Put $t^{2}+9$= s Then, $\frac{ds}{dt}= 2t$ As y= $s^{1/2}$, we have $\frac{dy}{ds}= \frac{1}{2}×s^{-\frac{1}{2}}= \frac{1}{2√s}$ Substituting the value of s, we get $\frac{dy}{ds}= \frac{1}{2√(t^{2}+9)}$ According to the chain rule, $\frac{dy}{dt}=\frac{dy}{ds}.\frac{ds}{dt}$= $\frac{1}{2√(t^{2}+9)}2t$= $\frac{t}{\sqrt (t^{2}+9)}$
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