Question 683193
Find the slope of the tangent line of x^2/3 + y^2/3 = 4 at the point (-1,3sqrt3).
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{{{x^2 + y^2 = 12}}}
differentiate implicitly
2xdx + 2ydy = 0
dy/dx = -y/x (for all cirlces)
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The point is not on the circle, tho.