Answer
$-\dfrac{2e^{\cos (\ln x^2)}}{x}$
Work Step by Step
Here, $y'=-e^{\cos (\ln x^2)}(\dfrac{d (\ln x^2)}{dx}$
We need to apply the chain rule:
Then $y'=-e^{\cos (\ln x^2)} (\dfrac{1}{x^2}) (2x)$
or, $y'=-\dfrac{2e^{\cos (\ln x^2)}}{x}$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Practice Exercises - Page 344: 125
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