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...

ACHIEVE FOR CALCULUS 4 TERM >CSI CUSTOM<

21st Edition

ISBN: 9781319438333

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 3.8, Problem 64E

To determine

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 $(2, \frac{1}{2}), (- 4, \frac{- 1}{4})$

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Chapter 3.8, Problem 63E

Chapter 3.8, Problem 65E

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Chapter 3 Solutions

ACHIEVE FOR CALCULUS 4 TERM >CSI CUSTOM<

Ch. 3.1 - Prob. 1PQ Ch. 3.1 - Prob. 2PQ Ch. 3.1 - Prob. 3PQ Ch. 3.1 - Prob. 4PQ Ch. 3.1 - Prob. 5PQ Ch. 3.1 - Prob. 6PQ Ch. 3.1 - Prob. 7PQ Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E

Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E Ch. 3.1 - Prob. 11E Ch. 3.1 - Prob. 12E Ch. 3.1 - Prob. 13E Ch. 3.1 - Prob. 14E Ch. 3.1 - Prob. 15E Ch. 3.1 - Prob. 16E Ch. 3.1 - Prob. 17E Ch. 3.1 - Prob. 18E Ch. 3.1 - Prob. 19E Ch. 3.1 - Prob. 20E Ch. 3.1 - Prob. 21E Ch. 3.1 - Prob. 22E Ch. 3.1 - Prob. 23E Ch. 3.1 - Prob. 24E Ch. 3.1 - Prob. 25E Ch. 3.1 - Prob. 26E Ch. 3.1 - Prob. 27E Ch. 3.1 - Prob. 28E Ch. 3.1 - Prob. 29E Ch. 3.1 - Prob. 30E Ch. 3.1 - Prob. 31E Ch. 3.1 - Prob. 32E Ch. 3.1 - Prob. 33E Ch. 3.1 - Prob. 34E Ch. 3.1 - Prob. 35E Ch. 3.1 - Prob. 36E Ch. 3.1 - Prob. 37E Ch. 3.1 - Prob. 38E Ch. 3.1 - Prob. 39E Ch. 3.1 - Prob. 40E Ch. 3.1 - Prob. 41E Ch. 3.1 - Prob. 42E Ch. 3.1 - Prob. 43E Ch. 3.1 - Prob. 44E Ch. 3.1 - Prob. 45E Ch. 3.1 - Prob. 46E Ch. 3.1 - Prob. 47E Ch. 3.1 - Prob. 48E Ch. 3.1 - Prob. 49E Ch. 3.1 - Prob. 50E Ch. 3.1 - Prob. 51E Ch. 3.1 - Prob. 52E Ch. 3.1 - Prob. 53E Ch. 3.1 - 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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 )

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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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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 $(2, \frac{1}{2}), (- 4, \frac{- 1}{4})$