# Wierd implicit derivative

1. Nov 10, 2008

### thharrimw

when i took the derivative of X^2*Y^2-2x=6 i didn't get the same thing that i got when i took the derivative of y^2=(6+2x)/x^2 and all i did was rewrite the origional equation. why did rewriting the equation change my answer?

2. Nov 10, 2008

### Staff: Mentor

Could you show us your work? That would make it easier for us to help.

3. Nov 10, 2008

### thharrimw

not rewritting i gou
(x^2)(Y^2)-2x=3
(x^2)(2y)(y')+2x(y^2)-2=0
(x^2)(2y)(y')=2-(2x)(Y^2)
y'=(1-x(y^2))/((x^2)2y))

4. Nov 10, 2008

### thharrimw

when i rewrote it i got y^2=(3+2x)/(x^2)
and my derivitive was (x-3)/(y(x^3))

5. Nov 10, 2008

### Staff: Mentor

I'm not tracking your math. Are you differentiating the equations with respect to x? It would help if you used LaTex to be explicit about your differentiations...

$$\frac{dy}{dx} = etc.$$

http://en.wikipedia.org/wiki/Implicit_function

.

6. Nov 10, 2008

### HallsofIvy

Okay, $x^2y^2- 2x= 3 has derivative [itex]2xy^2+ 2x^2y y'- 2= 0 so [itex]2x^2y y'= 2- 2xy^2$ $y'= (2-2xy^2)/(2x^2y)$, the "2"s cancel and y'= (1- xy^2)/(x^2y). You seem to have lost the "2" in the numerator.

No, solving for y2 gives $x^2 y^2=6+ 2x$ so $y^2= (6+ 2x)/x^2$ The derivative of $y^2$, with respect to x, is 2y y' and the derivative of $(6+2x)/x^2$ using the quotient rule is $[(2)x^2- (6+2x)(2x)]/(x^4)= [-2x^2- 12x]/(x^4)= -(2x^2+ 12x)/(x^4)= -2(x+ 6)/(x^3)$. $2yy'= -2(x+6)/x^3$ gives $y'= -(x+1)/yx^3$.

Think those aren't the same? Replace the y2 in the numerator of the first derivative above with (2x+6)/x2 and simplify.

Last edited by a moderator: Nov 10, 2008
7. Nov 11, 2008

### thharrimw

$2yy'= -2(x+6)/x^3$ gives $y'= -(x+1)/yx^3$.

wouldn't that simplify to $y'= -(x+6)/yx^3$.

8. Nov 14, 2008

### JANm

When I rewrite this I get (-x-3)/(y*x^3). Could that be the problem