Implicit derivative problem has two algebraically unequal answers with the exact same graph (??)

Question

In my math class we were assigned to find the implicit derivative of

$\dfrac yx + \dfrac xy= 2x$

And most of the class found the answer to be

$\text{(student answer)} \to\dfrac{dy}{dx} = \dfrac{(y^3-x^2y)}{(xy^2 -x^3 -2x^2y^2)}$

But the teacher had the answer

$\text{(teacher answer)}\to\dfrac{dy}{dx} = \dfrac{(y^2-x)}{(y-2xy)}$

She says that she sees how all of her students got the their answers, but she didn't understand why both answers were valid even when they are not algebraically equivalent. She says that she suspects it has something to do with variables acting as zeros in the denominator of the original function.

Can anyone explain why this happens?

Answer 1

It's been a while since I did an implicit differentiation but I get.........

Rearranging..............$$\frac{y^2+x^2}{xy} = 2x$$ $$y^2 + x^2 = 2x^2y$$ $$2y\frac{dy}{dx} + 2x = 4xy + 2x^2\frac{dy}{dx}$$ $$2y\frac{dy}{dx} - 2x^2\frac{dy}{dx} = 4xy - 2x$$ $$\frac{dy}{dx}(2y - 2x^2) = 4xy -2x$$ $$\frac{dy}{dx} = \frac{4xy - 2x}{2y -2x^2}$$ $$\frac{dy}{dx}= \frac{2xy - x}{y - x^2}$$

I notice from the original equation, putting in an x value $>1$ yields two y values. The equation is not a function (two y values for one x value) so this may explain the differences in the derivative.

Answer 2

I'm not sure about how the two answers were reached: either you differentiate as is, getting $$ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} + \frac{1}{y} - \frac{x}{y^2} \frac{dy}{dx} - 2 = 0, $$ which can be rearranged to $$ \frac{dy}{dx} = \frac{2+y/x^2-1/y}{1/x-x/y^2} = \frac{2x^2y+y^2-x^2}{y(y^2-x^2)} $$ or you clear denominators to start with $$ x^2+y^2 - 2x^2y =0, $$ of which differentiation gives $$ 2x + 2y \frac{dy}{dx} - 4xy - 2x^2\frac{dy}{dx} = 0, $$ and can be rearranged into $$ \frac{dy}{dx} = \frac{2xy-x}{y-x^2}. $$ How are these two equivalent? It becomes slightly more apparent if we multiply numerator and denominator of the first one by the denominator of the original formula $xy$: $$ \frac{2+y/x^2-1/y}{1/x-x/y^2} = \frac{2xy + y^2/x - x}{y-x^2/y} = \frac{(2xy-x) + y(y/x)}{y-x^2 +x(x-x/y)} $$ But now we can use the original formula $x/y+y/x=2x$: $$ \frac{(2xy-x) + y(y/x)}{y-x^2 +x(x-x/y)} = \frac{(2xy-x) + y(2x-x/y)}{y-x^2 +x(-x+y/x)} = \frac{(2xy-x) + (2xy-x)}{y-x^2 +(y-x^2)} = \frac{2xy-x}{y-x^2}. $$

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What's actually going on here? Suppose we have $f(x,y)/g(x,y) = 0$. Differentiating gives $$ \frac{f_x}{g} - \frac{fg_x}{g^2} + \left( \frac{f_y}{g} - \frac{fg_y}{g^2} \right) \frac{dy}{dx}, $$ where $f_x$ is the partial derivative $\partial f/\partial x$ (i.e. the derivative holding $y$ constant), which rearranges to $$ \frac{dy}{dx} = \frac{-f_x+(f/g)g_x}{f_y-(f/g)g_y} = \frac{-gf_x+fg_x}{gf_y-fg_y}. $$ But $f=0$, so this cancels to $dy/dx = -f_x/f_y$ as it should be. This is obscured when we have actual expressions for $f$ and $g$, so that there can be cancellations between the two terms in the numerator and/or denominator, hence why it was helpful to take a step back from the fully simplified version.