find where the slope is defined..

**Morgan82** · September 23rd 2009, 04:39 PM

(x^2)y - x(y^2) = 4

I came out with...

y' = [(y^2) - 2xy] / [(x^2)-2xy)]

The answer in the back isnt this...
How do you do this?

**artvandalay11** · September 23rd 2009, 04:45 PM

Originally Posted by Morgan82

(x^2)y - x(y^2) = 4

I came out with...

y' = [(y^2) - 2xy] / [(x^2)-2xy)]

The answer in the back isnt this...
How do you do this?

$x^2y-xy^2=4$

So we have to differentiate implicitly,

$<br /> 2x(y)+x^2y'-1\cdot y^2-x(2y)y'=0<br />$

And solve for y' to get

$x^2y'-2xyy'=y^2-2xy$

$y'(x^2-2xy)=y^2-2xy$

$y'=\frac{y^2-2xy}{x^2-2xy}$

The slope is defined as long as the denominator doesn't equal 0.

It is not defined when x=0, it is not defined when $y=\frac{x}{2}$

**skeeter** · September 23rd 2009, 04:52 PM

Originally Posted by Morgan82

(x^2)y - x(y^2) = 4

I came out with...

y' = [(y^2) - 2xy] / [(x^2)-2xy)]

The answer in the back isnt this...
How do you do this?

I agree with you ... what "answer" was in the back of the book?

$x^2y' + 2xy - 2xy y' - y^2 = 0$

$x^2y' - 2xy y' = y^2 - 2xy$

$y'(x^2 - 2xy) = y^2 - 2xy$

$y' = \frac{y^2-2xy}{x^2-2xy}$

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