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: 24


$y'=\dfrac{\csc(x+1)^3}{2\sqrt x} -3\sqrt x \csc (x+1)^3 \cot (x+1)^3 (x+1)^2$

Work Step by Step

Given that $y=\sqrt x \csc(x+1)^3 $ Apply derivative rules of differentiation: $f(x)=p'(x)q(x)+p(x)q'(x)$ $y'=\dfrac{d}{dx}[\sqrt x \csc(x+1)^3]$ $=\dfrac{1}{2\sqrt x} \csc(x+1)^3+\sqrt x[-csc (x+1)^3 \cot (x+1)^3]\dfrac{d}{dx}(x+1)^3$ $=\dfrac{\csc(x+1)^3}{2\sqrt x} -\sqrt x \csc (x+1)^3 \cot (x+1)^3 \cdot 3(x+1)^2$ Hence, $y'=\dfrac{\csc(x+1)^3}{2\sqrt x} -3\sqrt x \csc (x+1)^3 \cot (x+1)^3 (x+1)^2$
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