# find where the slope is defined..

• Sep 23rd 2009, 04:39 PM
Morgan82
find where the slope is defined..
(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?
• Sep 23rd 2009, 04:45 PM
artvandalay11
Quote:

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?

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

So we have to differentiate implicitly,

$\displaystyle 2x(y)+x^2y'-1\cdot y^2-x(2y)y'=0$

And solve for y' to get

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

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

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

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 $\displaystyle y=\frac{x}{2}$
• Sep 23rd 2009, 04:52 PM
skeeter
Quote:

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?

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

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

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

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