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Use implicit differentiation to find an equation of the tangent line to the curve at the given point (2,4)?

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x^2+2xy-y^2+x=6

asked Nov 1, 2014 in PRECALCULUS by anonymous

1 Answer

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The curve equation is x2 + 2xy - y2 + x = 6

Differentiating on each side with respect to x .

2x + 2[x(dy/dx) + y] - 2y (dy/dx) + 1 = 0

2x + 2x (dy/dx) + 2y - 2y(dy/dx) + 1 = 0

dy/dx [2x - 2y] =  - 1 - 2x - 2y

dy/dx [2x - 2y] = - 1 - 2x - 2y

dy/dx = (- 1 - 2x - 2y)/(2x - 2y)

y' = (- 1 - 2x - 2y)/(2x - 2y)

 

Substitute the values (x , y ) = (2, 4) in above equation.

y' = [- 1 - 2(2) - 2(4)]/[2(2) - 2(4)]

y' = (- 1 - 4 - 8)/(4 - 8)

y' = 13/4

This is the slope of tangent line to the curve at (2, 4).

 

To find the tangent line equation, substitute the values of m = 13/4 and (x, y ) = (2, 4) in the slope intercept form of an equation y = mx + b.

4 = (13/4)2 + b

4 = (13/2) + b

b = 4 - (13/2)

b = ( 8 - 13)/2

b = - 5/2

Substitute m = 13/4 and b = 1/2 in y = mx + b.

Tangent line equation is y = (13x/4) - (5/2).

answered Nov 1, 2014 by david Expert

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