Find dy/dx

2x^2 y^2 = x^3 y^3

I got dy/dx = (3x^2 y^3 - 4xy^2) / (4x^2 y - 3x^3 y ^2) but the answer is apparently different. Can anyone help me with this question?

2 answers

2x^2 y^2 = x^3 y^3

I would first simply it to
2 = xy , where x, y ≠ 0
0 = x dy/dx + y
dy/dx = -y/x

use
www.desmos.com/calculator

to graph both xy = 2 and
2x^2 y^2 = x^3 y^3
on the same grid to show that they are the same
2x^2 y^2 = x^3 y^3
4xy^2 + 4x^2 yy' = 3x^2 y^3 + 3x^3 y^2 y'
y'(4x^2 y - 3x^3 y^2) = 3x^2 y^3 - 4xy^2
as you said. But maybe you should have started by dividing both sides by x^2 y^2 to get
xy = 2
y + xy' = 0
y' = -y/x

I'm sure that with enough algebra, you could manipulate your answer into this form -- give it a shot
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