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Hi can anyone please tell me where I goes wrong with this question:

Find $ \frac{dy}{dx} $ for the curves defines by this equation:

\begin{align} \frac{x}{y} + \frac{y}{x} = 1 \end{align}

Here is what I did:

\begin{align} &\frac{y-xy'}{y^2} + \frac{y'x - y}{x^2} =0 \\ &\frac{yx^2 - x^3 y'+y'y^2 x - y^3}{x^2 y^2} = 0 \\ &yx^2 + (-x^3 +y^2 x)y' -y^3 =0 \\ &\therefore y'= \frac{y^3 -yx^2}{-x^3 +y^2x} \end{align}

The answer say it should be: $ y' = \frac{y}{x} $ but I had no clue how to proceed from there.

Please help, Thanks.

abel
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Justin HT
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  • Welcome to math.stackexchange! Nice work explaining and typesetting your work. As to your question: You can cancel $y^2-x^2$ in your expression and then you are there. – Hans Engler Jan 17 '15 at 14:55
  • Oh wow I can't believe I didn't see it when I post the question! Thanks very much! Hans – Justin HT Jan 17 '15 at 14:56
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    If we write $u := \frac{x}{y}$, then the l.h.s. is $u + \frac{1}{u}$, but this does not take values in $(-2, 2)$ for real $u$, so the given relation is empty and there's nothing to compute. – Travis Willse Jan 17 '15 at 15:45
  • @Travis did you mean the equation is not a relation and thus should Not have a $ \frac{dy}{dx}$ ? – Justin HT Jan 17 '15 at 15:51
  • The equation defines a relation, but there are no $(x, y)$ that satisfy it, so the relation is empty, and hence there are no points at which to compute a derivative. – Travis Willse Jan 17 '15 at 15:56
  • @Travis now that's quite interesting, if no (x, y) satisfies the relation then what does the supposed "solution" $ y' = \frac{y}{x} $ represent? (if not derivative) – Justin HT Jan 17 '15 at 16:00
  • At least it is the correct formula when we replace the r.h.s. of the defining equation with a value the l.h.s. does achieve. – Travis Willse Jan 17 '15 at 16:05
  • @Travis I think this question is not intended to be interpreted geometrically. – Justin HT Jan 17 '15 at 16:07
  • @JustinHT So how should it be interpreted? That $x, y$ can be complex numbers? –  Jan 17 '15 at 20:19
  • @JustinHT The issue is not geometric, it's set-theoretic. – Travis Willse Jan 18 '15 at 04:18
  • @LeonAragones I'm not sure too, this question confused me a lot. – Justin HT Jan 19 '15 at 08:24
  • @Travis sorry I'm not familiar to with set theory. – Justin HT Jan 19 '15 at 08:27
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    @JustinHT It's quite possible the question's author didn't foresee this complication. Probably it's best just to treat it as though the quantity on the r.h.s. is, e.g., $2$, in which case the arguments in the answers still apply (and are not vacuous). – Travis Willse Jan 19 '15 at 08:31
  • @Travis yes I would agree. I think the author just intended it to be a practice question in implicit differentiation. – Justin HT Jan 19 '15 at 08:45

3 Answers3

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if you going to do implicit differentiation you might as well multiply by $xy$ to get $x^2 + y^2 = xy$ before differencing. now differencing gives you $2xdx + 2ydy = xdy + ydx.$ this can also be written as $$\dfrac{dy}{dx} = \dfrac{y - 2x}{2y-x} = \dfrac{y(y-2x)}{y(2y-x)}= \dfrac{y(y-2x)}{2y^2 - xy} = \dfrac{y(y-2x)}{2xy - 2x^2 - xy} = \dfrac{y(y-2x)}{x(y-2x)} = \dfrac{y}{x} \tag 1$$

in implicit differentiation you don't have a unique answer. you always have to carry the constraint $\dfrac{x}{y} + \dfrac{y}{x} = 1$ along or an equivalent one like $x^2 + y^2 = xy$ along with the solution. in the equality (1) any one of them could be an answer.

abel
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  • Hi abel can you explain how did you do this transformation: $$ \dfrac{y(y-2x)}{2y^2 - xy} = \dfrac{y(y-2x)}{2xy - 2x^2 - xy} $$ ? – Justin HT Jan 17 '15 at 15:21
  • @JustinHT, because i wanted an $y$ in the numerator and then the denominator better has a factor $y- 2x$ if we are going to match answer in the book. – abel Jan 17 '15 at 15:27
  • I meant how not why. – Justin HT Jan 17 '15 at 15:30
  • @JustinHT, i don't understand what you mean. we have $\frac{dy}{dx} = \frac{y-2x}{2y - x}$ i just multiplied the numerator and denominator of the right hand side by $y.$ – abel Jan 17 '15 at 15:32
  • yes that would turn $$ \dfrac{y - 2x}{2y-x}$$ to $$\dfrac{y(y-2x)}{y(2y-x)} $$ but how did you turn this $$\dfrac{y(y-2x)}{2y^2 - xy} $$ to $$\dfrac{y(y-2x)}{2xy - 2x^2 - xy}$$ – Justin HT Jan 17 '15 at 15:36
  • @JustinHT, because i used $y^2= xy - x^2$ – abel Jan 17 '15 at 15:37
  • oh I got it now Thanks! @abel – Justin HT Jan 17 '15 at 15:39
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Note that $y=y(x)$

So we have that $$\frac{x}{y(x)} + \frac{y(x)}{x} = 1$$

So,

$$\begin{align} 0 &= \frac{d}{dx}\left( \frac{x}{y(x)} + \frac{y(x)}{x}\right) \\ &=\frac{d}{dx}(x) \frac{1}{y(x)}+x \frac{d}{dx}\left(\frac{1}{y(x)}\right) \\ &= \frac{1}{y(x)}+ x \ \frac{-1}{y^2(x)} \frac{dy}{dx} \\ &= \frac{1}{y(x)}\left(1-\frac{x}{y(x)} \right)\frac{dy}{dx} \end{align}$$

Assuming $\frac{1}{y(x)} \neq 0$ then

$$0= \left(1-\frac{x}{y(x)} \right)\frac{dy}{dx} $$

$$-1= -\frac{x}{y(x)} \frac{dy}{dx} $$

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

The only thing I used is the product rule http://en.wikipedia.org/wiki/Product_rule

Trajan
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we get by the quotient and the chaine rule $$\frac{y-xy'}{y^2}+\frac{y'x-y}{x^2}=0$$ multiplying by $x^2y^2$ we obtain $$x^2y-x^3y'+y'y^2x-y^3=0$$ solving for $y'$ we get $$y'=\frac{x^2y-y^3}{x^3-xy^2}$$ if $x^3-y^2x\ne 0$

Dr. Sonnhard Graubner
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