University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 25

Answer

$y'=-10x \csc^2 x^2$

Work Step by Step

Given that $y=5 \cot x^2$ Apply derivative rules of differentiation: $f(x)=p'(x)q(x)+p(x)q'(x)$ $y'=\dfrac{d}{dx}[5 \cot x^2]$ $=-5 \csc^2 x^2 \dfrac{d}{dx}[x^2]$ $=-5 \csc^2 x^2 (2x)$ Hence, $y'=-10x \csc^2 x^2$
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