Calculus 10th Edition

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

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 373: 79

Answer

$$y = - x + \sqrt 2 $$

Work Step by Step

$$\eqalign{ & \arcsin x + \arcsin y = \frac{\pi }{2} \cr & {\text{Using implicit differentiation}} \cr & \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{{y'}}{{\sqrt {1 - {y^2}} }} = 0 \cr & {\text{Solve for }}y' \cr & \frac{{y'}}{{\sqrt {1 - {y^2}} }} = - \frac{1}{{\sqrt {1 - {x^2}} }} \cr & y' = - \frac{{\sqrt {1 - {y^2}} }}{{\sqrt {1 - {x^2}} }} \cr & {\text{Calculate }}y'{\text{ at the given point }}\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right) \cr & y' = - \frac{{\sqrt {1 - {{\left( {\sqrt 2 /2} \right)}^2}} }}{{\sqrt {1 - {{\left( {\sqrt 2 /2} \right)}^2}} }} \cr & y' = - 1 \cr & {\text{The slope at the given point is }}m = - 1 \cr & {\text{The equation of the tangent line at }}\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right){\text{ is}} \cr & y - \frac{{\sqrt 2 }}{2} = - \left( {x - \frac{{\sqrt 2 }}{2}} \right) \cr & y - \frac{{\sqrt 2 }}{2} = - x + \frac{{\sqrt 2 }}{2} \cr & y = - x + \sqrt 2 \cr} $$
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