Answer
$-2x \sin (x^2-1)$
Work Step by Step
We have $f(x)=\cos (x^2-1)$
We differentiate both sides with respect to $x$.
$f'(x)= \dfrac{d}{dx}[\cos (x^2-1)]$
or, $f'(x)=-\sin (x^2-1) \dfrac{d}{dx}[(x^2-1)]=-2x \sin (x^2-1)$
Finite Math and Applied Calculus (6th Edition)
by Waner, Stefan; Costenoble, Steven
Published by
Brooks Cole
ISBN 10:
1133607705
ISBN 13:
978-1-13360-770-0
Chapter 16 - Review - Review Exercises - Page 1181: 9
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