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

Question

Isaiah Allen · Accepted Answer

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.)

Scarlett Allison · Answer

undefined To differentiate the implicit equation $x^2 + y^2 = 2xy$, follow these steps:

1. Implicitly differentiate both sides of the equation with respect to $x$.
2. Use the chain rule and product rule where necessary.
3. Solve for $\frac{{dy}}{{dx}}$ to find the derivative of $y$ with respect to $x$.

Differentiating both sides with respect to $x$:

$$ \frac{{d}}{{dx}}(x^2 + y^2) = \frac{{d}}{{dx}}(2xy) $$

Using the chain rule and product rule:

$$ 2x + 2y\frac{{dy}}{{dx}} = 2y + 2x\frac{{dy}}{{dx}} $$

Rearranging terms and solving for $\frac{{dy}}{{dx}}$:

$$ 2y\frac{{dy}}{{dx}} - 2x\frac{{dy}}{{dx}} = 2x - 2y $$
$$ \frac{{dy}}{{dx}}(2y - 2x) = 2x - 2y $$
$$ \frac{{dy}}{{dx}} = \frac{{2x - 2y}}{{2y - 2x}} $$

Simplify:

$$ \frac{{dy}}{{dx}} = \frac{{x - y}}{{y - x}} $$

Therefore, the derivative of $y$ with respect to $x$ for the given implicit equation is $\frac{{x - y}}{{y - x}}$.