Explanation Derivative of a function y = f ( x ) tells us the rate at which value of y changes in the function with respect to the change in value of x. This change is represented as slope of the line. Equation of line is given by y − y 1 = ( d y d x ) x = x 1 y = y 1 ( x − x 1 ) Given: The given equation is x 2 y − 2 x + 8 y = 2 . Formula used: We shall use the chain rule and d d x ( x n ) = n x n − 1 Calculations: we are given with the equation x 2 y − 2 x + 8 y = 2 Differentiate both sides with respect to x we get d ( x 2 y − 2 x + 8 y = 2 ) d x 2 x y + x 2 y ' − 2 + 8 y ' = 0 y ' ( x 2 + 8 ) = 2 − 2 x y y ' = 2 − 2 x y x 2 + 8 Where d y d x = y ' Since the tangent is horizontal

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 3.8, Problem 40E

To determine

The equation of tangent to the line at the indicated point (x,y), x2y−2x+8y=2 horizontal then xy=1 .then substitute y=x−1 in x2y−2x+8y=2 to show that tangent line is horizontal at the point (2,12),(−4,−14)

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Chapter 3.8, Problem 39E

Chapter 3.8, Problem 41E

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The equation of tangent to the line at the indicated point ( x , y ) , x 2 y − 2 x + 8 y = 2 horizontal then x y = 1 .then substitute y = x − 1 in x 2 y − 2 x + 8 y = 2 to show that tangent line is horizontal at the point ( 2 , 1 2 ) , ( − 4 , − 1 4 )

Solution Summary: The author explains that the equation of tangent to the line at the indicated point (x,y) , x2y-2x+8y=2 horizontal.

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The equation of tangent to the line at the indicated point (x,y), x2y−2x+8y=2 horizontal then xy=1 .then substitute y=x−1 in x2y−2x+8y=2 to show that tangent line is horizontal at the point (2,12),(−4,−14)

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