# Implicit diff on $\frac{2x}{y} - \frac{3y}{x} = 8$

Implicit diff on $\frac{2x}{y} - \frac{3y}{x} = 8$

I'm doing this:

$\frac{2x^2 - 3y^2}{xy} = 8$

$2x^2 - 3y^2 = 8xy$

$2x^2 - 3y^2 - 8xy = 0$

Derive

$4x dx - 6y dy - 8( ydx + xdy ) = 0$

$dx( 4x - 8y ) - dy ( 6y + 8x ) = 0$

$\frac{dy}{dx} = \frac{4x - 8y}{ 8x + 6y }$

But if you use the quotient rule from the beginning, you get

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

Can someone point out to me where I've gone wrong?

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Note that you are introducing a lot of "false positives". The function implicitly defined by $$\frac{2x}{y} - \frac{3y}{x} = 8$$ is not the same as the function implicitly defined by $$2x^2 - 3y^2 = 8xy.$$ Why not? Well, the second function includes $y(0)=0$ (that is, the point $(0,0)$), but the first one is not defined for $x=0$ or $y=0$! Now, it is true that any point on the second curve in which $x\neq 0$ and $y\neq 0$ will be on your original curve, and that the only point in the second curve not on the orginal one is $(0,0)$, so you can make some progress.

These are things you must be careful with when you perform that kind of easy manipulations with implicit functions.

Now, what you seem to be forgetting is that the values of $x$ and of $y$ are connected by the equation $\frac{2x}{y}-\frac{3y}{x} = 8$. So it is entirely possible that the formula $\frac{y}{x}$ will give the exact same values as $$\frac{4x-8y}{8x+6y} = \frac{2(x-y)}{4x+3y}$$ for any $(x,y)$ that satisfies $\frac{2x}{y}-\frac{3y}{x}=8$. Let's see if that's the case.

We must assume that $x\neq 0$ and that $y\neq 0$; in addition, your formula has the potential unpleasant side effect of also prohibiting points on the line $4x+3y=0$. Are there any points on that line that are on the original curve? If $y = -\frac{4}{3}x$, then plugging into the left hand side of your original equation we have $$-\frac{6x}{4x} + \frac{4x}{x} = -\frac{6}{4}+4 \neq 8$$ so, phew!, none of the points your formula excludes are on the original graph.

Now, some manipulation is in order. Since $x\neq 0$, $y\neq 0$, and $4x+3y\neq 0$, we have: \begin{align*} \frac{y}{x} = \frac{2x-4y}{4x+3y} &\Longleftrightarrow y(4x+3y) = (2x-4y)x\\ &\Longleftrightarrow 4xy + 3y^2 = 2x^2 - 4xy\\ &\Longleftrightarrow 8xy = 2x^2 - 3y^2\\ &\Longleftrightarrow 8 = \frac{2x}{y} - \frac{3y}{x}. \end{align*} So not only are the two expressions equal on the curve you are working on, they are equal only on the curve you are working on. So the two answers are equivalent, though yours is by far more complicated.

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Did you try checking the possibility that $\displaystyle \frac{4x-8y}{8x+6y} = \frac{y}{x}$?

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It can't be.. I tried checking x=2, y=1. Then RS=$\frac{1}{2}$, LS=0. – bobobobo Feb 2 '11 at 7:21
@bobo: They are functions aren't they? Why are you trying to set $x$ to a constant? In any case, if you set them equal you get $2x^2 - 3y^2 = 8xy$ which is consistent with what you started out with. – Aryabhata Feb 2 '11 at 7:25
@bobobobo : if x=2 and y=1, 2x/y - 3y/x is not 8, so you're not on the curve anyway. Taking such particular values for x and y is meaningless. – mercio Feb 2 '11 at 10:43

I seriously doubt the jump from equation before the word Derive to equation after the word in your question. There are definitely some voluntaristic differentiations present.

Implicit differentiation of implicit function $F(x,y)=0$ is carried out in the following way:

$\frac{dy}{dx} = - \frac{\partial F/ \partial x}{\partial F/ \partial y}$

We've got $F(x,y)= \frac{2x}{y} - \frac{3y}{x} - 8$

Then $\frac{\partial F}{\partial x} = \frac{2}{y} + \frac{3y}{x^2} = \frac{2x^2+3y^2}{x^2y}$, and $\frac{\partial F}{\partial y} = -\frac{2x}{y^2} - \frac{3}{x} = -\frac{2x^2+3y^2}{xy^2}$.

Dividing finally gives $\frac{dy}{dx}= -\frac{\frac{2x^2+3y^2}{x^2y}}{-\frac{2x^2+3y^2}{xy^2}} = {\frac{2x^2+3y^2}{x^2y}} \cdot \frac{xy^2}{2x^2+3y^2} = \frac{y}{x}$.

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You don't need derivation to solve this equation. $x \neq 0$, so $3y^2 + 8xy - 2x^2 = 0$ is equivalent to $3\frac{y}{x}^2 + 8\frac{y}{x} - 2 = 0$, which is simply a quadratic equation.

Then you get $y = k_ix$, where $k_1$, $k_2$ are the roots of the equation. Note that it does imply that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x} = k_i$ :)

EDIT: if either of $k_i$ is zero, the corresponding function will obviously not be a solution to the original equation.

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Nobody is trying to "solve the equation" (that is, nobody is trying to write $y$ as an explicit function of $x$); the intent is to find $\frac{dy}{dx}$ directly, without having to go through the process of trying to obtain an explicit formula for $y$ as a function of $x$ (because you don't want to have to deal with different formulas when trying to find $y'$ depending on the value of $x$, and because you don't want to have to to derivatives of square roots when you can avoide them). – Arturo Magidin Feb 2 '11 at 20:14
Oh, OK then ^_^ – Alexei Averchenko Feb 2 '11 at 20:17