Math
posted by lana .
Suppose that x and y are related by the given equation and use implicit differentiation to determine dy/dx.
x^2y + y^2x = 3

Math 
Count Iblis
d[x^2y + y^2x] = 0 >
2xydx + x^2dy + 2xydy + y^2dx = 0 >
(2xy + y^2)dx + (2xy + x^2) dy = 0 >
dy/dx = (2xy + y^2)/(2xy + x^2) 
Math 
Reiny
use the product rule for each term on the left side
x^2(dy/dx) + y(2x)(dx/dx) + y^2(dx/dx) + x(2y)(dy/dx) = 0
x^2(dy/dx) + 2xy + y^2 + 2xy(dy/dx) = 0
dy/dx is a common factor, so ...
dy/dx(x^2 + 2xy) = 2xy  y^2
dy/dx = (2xy  y^2)/(x^2 + 2xy)
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