Explanation Given: The function is ln ( x 2 + y 2 ) = ln 5 x + y x − 2 and the point is ( 1 , 2 ) . Approach: Find the slop of curve and substitute the values of given point in the equation of line to get the solution. Calculation: Consider the equation. ln ( x 2 + y 2 ) = ln 5 x + y x − 2 ... (1) Differentiate Equation (1) with respect to x . ( 1 x 2 + y 2 ) ( 2 y + 2 x d y d x ) = 5 5 x + x d y d x − y x 2 − 0 ( 1 x 2 + y 2 ) ( 2 y + 2 x d y d x ) = 1 x + 1 x d y d x − y x 2 d y d x = 1 x − y x 2 − 2 x x 2 + y 2 2 y x 2 + y 2 − 1 x

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ISBN: 9780321964038

Author: GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher: Pearson Addison Wesley,

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Chapter 6.3, Problem 28E

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The equation of the tangent line at the given curve ln(x2+y2)=ln5x+yx−2 and the point is (1,2).

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Chapter 6.3, Problem 27E

Chapter 6.3, Problem 29E

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The equation of the tangent line at the given curve ln ( x 2 + y 2 ) = ln 5 x + y x − 2 and the point is ( 1 , 2 ) .

Solution Summary: The author explains that the equation of the tangent line at the given point on the curve is y=7x-5.

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To find:

The equation of the tangent line at the given curve ln(x2+y2)=ln5x+yx−2 and the point is (1,2).

Want to see the full answer?

Chapter 6 Solutions

The equation of the tangent line at the given curve ln ( x 2 + y 2 ) = ln 5 x + y x − 2 and the point is ( 1 , 2 ) .

Solution Summary: The author explains that the equation of the tangent line at the given point on the curve is y=7x-5.

Videos

To find: The equation of the tangent line at the given curve ln(x2+y2)=ln5x+yx−2 and the point is (1,2).

Want to see the full answer?

Chapter 6 Solutions

To find:

The equation of the tangent line at the given curve ln(x2+y2)=ln5x+yx−2 and the point is (1,2).