How do you differentiate #x^2+y^2=2xy#?

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Answer 1 · 2015-07-16T14:35:10

Saikiran Reddy and Kwasi F. give excellent solutions to this. The answer, #dy/dx =1# might make us think about the question a bit.

For #x^2+y^2=2xy#, we get (by differentiating implicitly), #dy/dx =1#.

That's the same as the derivative of a linear function with slope, #1#. Hmmmmm. Let's see:

If we have
#x^2+y^2=2xy#

The we must also have:
#x^2-2xy +y^2=0#

Factoring gets us:
#(x-y)^2 = 0#

And the only way for that to happen is to have:
#x-y=0#

So #y=x#

and #dy/dx =1#.
(Which we already knew by differentiating, but this may be of interest as well.)