1. Mar 7, 2010

### benhou

1. The problem statement, all variables and given/known data

Could anyone explain that I got two different answers for this question: find $$dy/dx$$ of $$\frac{x}{x+y}-\frac{y}{x}=4$$

2. Relevant equations

3. The attempt at a solution

1. using quotient rule:
$$\frac{x+y-(1+dy/dx)x}{(x+y)^{2}}-\frac{x\frac{dy}{dx}-y}{x^{2}}=0$$
$$\frac{y}{(x+y)^{2}}-\frac{x}{(x+y)^{2}}\frac{dy}{dx}+\frac{y}{x^{2}}-\frac{1}{x}\frac{dy}{dx}=0$$
$$(\frac{x}{(x+y)^{2}}+\frac{1}{x})\frac{dy}{dx}=\frac{y}{(x+y)^{2}}+\frac{y}{x^{2}}$$
$$x(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})\frac{dy}{dx}=y(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})$$
$$\frac{dy}{dx}=y/x$$

2. simplify using common denominator before taking derivative:
$$\frac{x^{2}}{x^{2}+xy}-\frac{xy+y^{2}}{x^{2}+xy}=4$$
$$x^{2}-xy-y^{2}=4x^{2}+4xy$$
$$-y^{2}=3x^{2}+5xy$$
$$-2y\frac{dy}{dx}=6x+5y+5x\frac{dy}{dx}$$
$$\frac{dy}{dx}=-\frac{6x+5y}{2y+5x}$$

2. Mar 7, 2010

### Dick

They aren't really different. Pick a point on the original curve like y=1, x=(sqrt(13)-5)/6. Substitute it into both of your expressions for dy/dx. You'll get the same thing. Coincidence?

3. Mar 7, 2010

### Dick

Or just try it with a braindead simple example. Take y/x=1. Do it one way and you get dy/dx=y/x, do it the other way and you get dy/dx=1. Which one is right?

4. Mar 7, 2010

### benhou

I have tried a few examples, and they all came out to be the same. Wow, I am so amazed that two equations that look so different can be the same thing.
I also tried to let the two equations equal:
1. $$\frac{dy}{dx}=y/x$$
2. $$\frac{dy}{dx}=-\frac{6x+5y}{2y+5x}$$
Simplifying it, I got
$$y/x= -\frac{6x+5y}{2y+5x}$$
$$2y^{2}+5xy=-6x^{2}-5xy$$
$$6x^{2}+10xy+2y^{2}=0$$
$$3x^{2}+5xy+y^{2}=0$$

Which is the same as the original equation, shown in the process using the second method, the one says "simplifying using common denominator".

5. Mar 7, 2010

### benhou

Ya, thank you so much. What I had always been doing were trying disprove the first method, the one saying "using quotient rule", because of the step following this:
$$x(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})\frac{dy}{dx }=y(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})$$

6. Mar 7, 2010

### Dick

Just to be clear. y/x and -(6x+5y)/(2y+5x) aren't the same expression. But you can prove they are equal if you combine them with x/(x+y)-y/x=4. Did you see post 3? It's a really simple example of the same thing.

7. Mar 7, 2010

### benhou

Ya, I know they are not the same thing. They are just related to the original equation.

By post three, you mean this?
I didn't understand it. Were you trying to tell me that all three of these equations relate to each other?

8. Mar 7, 2010

### Dick

Ok, (y/x)=1. (y/x)'=y'/x-y/x^2=0. Solve for y'=y/x. Or reduce the original equation and get y=x. So y'=1. y'=1 and y'=y/x LOOK different. Until you remember the original equation was y/x=1. It's the same sort of difference you are looking at. Just a lot more obvious.

9. Mar 7, 2010

### benhou

I was confused about what you meant by "do it". I thought you wanted me to plug that equation into the ones I was doing. Now it's clear. Thanks a lot.