Explanation Given Information: The provided equation of the curve is, x ln y + 2 x y = 2 and point is ( 1 , 1 ) . Formula used: The slope of tangent to a curve y = f ( x ) at point ( x , y ) is given by the derivative of the curve at that point. The equation of a line passing through points ( x 1 , y 1 ) and slope m is given by, y − y 1 = m ( x − x 1 ) Calculation: Consider the provided equation of curve, x ln y + 2 x y = 2 Now, find the derivative d y d x from x ln y + 2 x y = 2 by taking the derivative term by term on both sides of the equation as, d d x ( x ln y ) + d d x ( 2 x y ) = d d x ( 2 ) x d y d x + ln y d x d x + 2 x d y d x + 2 y d x d x = 0 x y ′ y + ln y + 2 x y ′ + 2 y = 0 Solve for d y d x as, y ′ ( x y + 2 x ) = − ln y − 2 y y ′ = − y ln y + 2 y 2 x + 2 x y Now, the provided point is ( 1 , 1 )

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Author: HARSHBARGER

Publisher: CENGAGE L

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Chapter 11.3, Problem 42E

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To calculate: The equation of the tangent line to the curve xlny+2xy=2 at (1,1).

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Chapter 11.3, Problem 41E

Chapter 11.3, Problem 43E

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Chapter 11 Solutions

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Ch. 11.1 - 1. Ch. 11.1 - 2. If Ch. 11.1 - Prob. 3CP Ch. 11.1 - 4. Find . Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Find the derivatives of the functions in Problems...

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The equation of the tangent line to the curve x ln y + 2 x y = 2 at ( 1 , 1 ) .

Solution Summary: The author calculates the equation of the tangent line to the curve, xmathrmlny+2xy=2, and the slope of a line passing through points.

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To calculate: The equation of the tangent line to the curve xlny+2xy=2 at (1,1).

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Chapter 11 Solutions