Explanation Given: The point P = ( 0.5 , 1 ) on the curve y 5 + y − 2 x = 1 . Formula used: The formula of linearization L ( x ) , centered at x = a is: L ( x ) = f ′ ( a ) ( x − a ) + f ( a ) Calculation: The curve is y 5 + y − 2 x = 1 . Differentiate the function with respect to x . 5 y 5 − 1 d y d x + d y d x − 2 = 0 5 y 4 d y d x + d y d x − 2 = 0 Simplify the above equation. 5 y 4 d y d x + d y d x − 2 = 0 5 y 4 d y d x + d y d x = 2 ( 5 y 4 + 1 ) d y d x = 2 d y d x = 2 5 y 4 + 1 Now, find the value of d y d x at point P = ( 0.5 , 1 )

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

4th Edition

ISBN: 9781319312886

Author: Rogawski, Jon, Adams, Colin, FRANZOSA, Robert

Publisher: W. H. Freeman

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Chapter 4.1, Problem 70E

To determine

Find dydx. Show that the linearization at P. Use L(x) to estimate the y− coordinate of the point on the curve where x=0.55.

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Chapter 4.1, Problem 69E

Chapter 4.1, Problem 71E

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Chapter 4 Solutions

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

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Find d y d x . Show that the linearization at P . Use L ( x ) to estimate the y − coordinate of the point on the curve where x = 0.55 .

Solution Summary: The author shows the linearization at P. The value of y(0.55) is 0.9833.

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Find dydx. Show that the linearization at P. Use L(x) to estimate the y− coordinate of the point on the curve where x=0.55.

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