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March 4, 2015

March 4, 2015

Posted by **jump** on Sunday, September 30, 2012 at 2:11pm.

- calculus -
**Reiny**, Sunday, September 30, 2012 at 3:34pmusing implicit derivatives :

x^3(2y)dy/dx + y^2 (3x^2) + 2 - 5 dy/dx = 0

dy/dx(2x^3 y - 5) = -3x^2 y^2 - 2

dy/dx = (-3x^2y^2 - 2)/(2x^3y - 5)

at the point (1,1)

dy/dx = (-3 - 2)/(2-5) = 5/3

now you have the slope of the tangent, and a point on that tangent.

Use the method you commonly use to find the equation of that straight line

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