Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 16

Answer

$\dfrac{dy}{dx}=-\dfrac{y^2}{{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}}.$

Work Step by Step

$\dfrac{d}{dx}(x)=\dfrac{d}{dx}(\sec{\dfrac{1}{y}})\rightarrow$ Using the Chain Rule with $u=y^{-1}\rightarrow\dfrac{du}{dx}=-\dfrac{1}{y^2}\rightarrow$ $\dfrac{d}{dx}(\sec{\dfrac{1}{y}})=\dfrac{dy}{dx}(-\dfrac{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}{y^2})\rightarrow$ $1=\dfrac{dy}{dx}(-\dfrac{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}{y^2})\rightarrow$ $\dfrac{dy}{dx}=-\dfrac{y^2}{{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}}.$
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