# Homework Help: Superior implicit differentiation, prove answer.

1. Sep 28, 2010

### Sakha

1. The problem statement, all variables and given/known data
$$2XY = Y^2 prove that y''2 = \frac{y^{2}-2xy}{(y-x)^3}$$
EDIT: Sorry, don't know how to insert a space in Latex.
2. Relevant equations

3. The attempt at a solution
$$2y+2x \frac{dy}{dx} = 2y \frac{dy}{dx}$$
$$\frac{dy}{dx}(2y-2x) = 2y$$
$$\frac{dy}{dx}= \frac{y}{y-x}$$
$$\frac{d^2y}{dx^2}=\frac{\frac{dy}{dx}(y-x)-(\frac{dy}{dx}-1)y}{(y-x)^2}$$
$$\frac{d^2y}{dx^2}= \frac{y^2}{(y-x)^3}$$

Is my work correct? If yes, then the question itself is wrong.
In the original equation (2XY = Y2) does it matters that the letters are caps?

2. Sep 28, 2010

### Dick

Your work is correct up to the last line. But the given solution is the correct one. Looks like something went wrong when you substituted your expression for dy/dx in and simplified.

3. Sep 28, 2010

### Sakha

Plugging in gives me:
((y/y-x)(y-x)-((y/y-x)-1)y)/((y-x)2)
Simplifiying:
(y-(y2/(y-x))-y))/((y-x)2)
(-y2/(y-x))/((y-x)2)

Different than my first try (now it's negative), but still I don't know how to the the -2xy in there.

4. Sep 28, 2010

### Dick

In y-(y^2/(y-x)-y) the y's don't cancel. It's minus of a minus.

5. Sep 28, 2010

### Sakha

Got it, my mistake was that I moved the (y-x) that was in the numerator as a denumerator before adding fractions. And the wrong minus of a minus that you pointed.
Thanks.