f'(x) =
2x +2y + 2xdy/dx - 2ydy/dx + 1 = 0
2xdy/dx - 2ydy/dx = -2x - 2y -1
dy/dx = (-2x - 2y - 1)/ (2x - 2y)
the tangent passes through (2,4)
m = dy/dx = (-2(2) - 2(4) -1 )
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(2(2) - 2(4))
m = 13/4
the equation of the tangent line through (2,4);
y = mx + c
4 = 13/4(2) + c
c = -5/2
y = 13/4x - 5/2
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x^2+2xy-y^2+x=6 , (2,4) (hyperbola)
2 answers
thanks for the help.