(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 $\displaystyle y=\frac{x}{2}$