# calculus

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what is the equation of the tangent line to the curve x^3 + 2y^2 + 3xy = 6 at the point (1,1)?

• calculus -

3x^2 + 4y dy/dx + 3xdy/dx + 3y = 0
dy/dx(4y+3x) = -3x^2 - 3y
dy/dx = (-3x^2 - 3y)/(4y+3x)

at (1,1) dy/dx = (-3-3)/(4+3) = -6/7

y-1 = (-6/7)(x-1
times 7
7y - 7 = -6(x-1)
7y - 7 = -6x + 6
6x + 7y = 13

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