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: 60


$$ y' =\frac{t}{\sqrt{t^2-9}}\sec\sqrt{t^2-9}\tan\sqrt{t^2-9}$$

Work Step by Step

Recall that $(\sec x)'=\sec x\tan x$. Recall that $(x^n)'=nx^{n-1}$ Since $ y=\sec\sqrt{t^2-9}$, by using the chain rule, the derivative $ y'$ is given by $$ y'=\sec\sqrt{t^2-9}\tan\sqrt{t^2-9} (\sqrt{t^2-9})'\\ =\frac{2t}{2\sqrt{t^2-9}}\sec\sqrt{t^2-9}\tan\sqrt{t^2-9}\\ =\frac{t}{\sqrt{t^2-9}}\sec\sqrt{t^2-9}\tan\sqrt{t^2-9}.$$
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