Why Does Rewriting an Equation Change the Implicit Derivative Result?

[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 original equation. why did rewriting the equation change my answer?

 

[berkeman]

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 original equation. why did rewriting the equation change my answer?

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

 

[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))

 

[thharrimw]

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

 

[berkeman]

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

.

 

[HallsofIvy]

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))

Okay, [itex]x^2y^2- 2x= 3 has derivative [itex]2xy^2+ 2x^2y y'- 2= 0 so $2x^2y y'= 2- 2xy^2$ $y'= (2-2xy^2)/(2x^2y)$, the "2"s cancel and y'= (1- xy^2)/(x^2y). You seem to have lost the "2" in the numerator.<br /> <br /> <blockquote data-attributes="" data-quote="thharrimw" data-source="post: 1951521" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> thharrimw said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> when i rewrote it i got y^2=(3+2x)/(x^2) and my derivative was (x-3)/(y(x^3)) </div> </div> </blockquote> 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 $y'= -(x+1)/yx^3$.<br /> <br /> Think those aren't the same? Replace the y² in the numerator of the first derivative above with (2x+6)/x² and simplify.[/itex][/itex]

 

[thharrimw]

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

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

 

[JANm]

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

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