Question
Asked Oct 29, 2019
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 Find an equation of the line tangent to the curve defined by x^3+2xy+y^3=4 at the point (1,1).

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Expert Answer

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Step 1

To find the equation of the tangent line at the point (1, 1) to the curve given below.

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-2xy = 4

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Step 2

Formula used: Equation of tangent line to a curve at a point is given by

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y mx-x) dy at (x.y) dx Here m

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Step 3

First find the derivative of y with respect to x by implicit...

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Differentiate both sided of the equation with respect to x d d d x+2(y) ) 0 = dx dy dy 3x2 yx +3y2 dx +31,2 dx dx 3x22x 0 dy -3x? - 2у dx 2x +3y =- dx (2x+3y2)3x-2y dx dy -3x-2y 2x+3y dx

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Tagged in

Math

Calculus

Derivative