Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.6 Implicit Differentiation - 2.6 Exercises - Page 167: 37

Answer

$y^{''}=\large\frac{\cos^2 y\;\cos x+\sin^2 x\;\sin y}{\cos^3 y}$

Work Step by Step

$\sin y+\cos x=1$ ____(1) Differentiating (1) implicitly with respect to $x$ $\cos y\; \cdot y^{'}-\sin x=0$ Solving for $y^{'}$ $y^{'}=\large\frac{\sin x}{\cos y}$ ____(2) Differentiating (2) implicitly with respect to $x$ $y^{''}=\large\frac{d}{dx}\left(\frac{\sin x}{\cos y}\right)$ $y^{''}=\large\frac{\cos y\;(\sin x)^{'}-\sin x\;(\cos y)^{'}}{\cos^2 y}$ $y^{''}=\large\frac{\cos y\;\cos x+\sin x\;\sin y\; \frac{dy}{dx}}{\cos^2 y}$ From (2) $y^{''}=\large\frac{\cos y\;\cos x+\sin x\;\sin y\; \left(\large\frac{\sin x}{\cos y}\right)}{\cos^2 y}$ $y^{''}=\large\frac{\cos^2 y\;\cos x+\sin^2 x\;\sin y}{\cos^3 y}$ Hence $y^{''}=\large\frac{\cos^2 y\;\cos x+\sin^2 x\;\sin y}{\cos^3 y}$.
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