• Oct 18th 2010, 09:04 PM
downthesun01
Find $\frac{\partial y}{\partial x}$ at the point (1,-1) for $3xy^{2}+5x^{2}y+4x=0$

So I came up with:
$(-3y^{2}\frac{\partial x}{\partial x}-6xy\frac{\partial y}{\partial x})+(10xy\frac{\partial x}{\partial x}+5x^{2}\frac{\partial y}{\partial x}+4\frac{\partial x}{\partial x})=0$

After cleaning it up a bit, I got:

$\frac{\partial y}{\partial x}(-6xy+5x^{2})=3 y^{2}-10xy-4$

$=>\frac{\partial y}{\partial x}=\frac{(3 y^{2}-10xy-4)}{(-6xy+5x^{2})}$

So:

$\frac{(3 y^{2}-10xy-4)}{(-6xy+5x^{2})}|_{(1,-1)}$

$=>\frac{(3 (-1)^{2}-10(1)(-1)-4)}{(-6(1)(-1)+5(1)^{2})}=\frac{9}{11}$

Can someone please double check this for me? Thanks
• Oct 19th 2010, 03:45 AM
HallsofIvy
You should not have a "-" on that $3y^2$ in the first line so the entire numerator in your derivative should be negative.

Also, I do not see why you call this a "partial" derivative and use $\frac{\partial y}{\partial x}$ rather than $\frac{dy}{dx}$. There does not appear to be any other independent variable.