Superior implicit differentiation, prove answer.

[Sakha]

Homework Statement

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.

Homework Equations

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?

 

[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.

 

[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.

 

[Dick]

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.

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

 

[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.