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: 12


$y'=\dfrac{-2 \cos x}{ \sin^3 x}+\dfrac{2 \cos x}{\sin^2 x}$

Work Step by Step

Given that $y=\dfrac{1}{ \sin^2 x}-\dfrac{2}{\sin x}$ Re-write the given equation as: $y= \sin^{-2} x-2\sin^{-1} x$ Apply derivative rules of differentiation: $f(x)=p'(x)q(x)+p(x)q'(x)$ $y'=\frac{d}{dx}[ \sin^{-2} x-2\sin^{-1} x]$ $= (-2)\sin^{-3} x \cos x+2\sin^{-2} x \cos x$ Hence, $y'=\dfrac{-2 \cos x}{ \sin^3 x}+\dfrac{2 \cos x}{\sin^2 x}$
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