# [SOLVED] tangent line to the graph

• Nov 4th 2007, 01:15 PM
simply07
[SOLVED] tangent line to the graph
Hey... i have no idea how to work with the y's in the equation. '
The question is:
Determine the equation of the tangent line to the graph of the relation with equation xy^2+x^2y=12 at the point where x=3.

Please any help would be greatly appreciated... thanks
• Nov 4th 2007, 01:54 PM
galactus
When it's difficult to solve for y, then we can use implicit differentiation.

$xy^{2}+x^{2}y=12$

$2xy\frac{dy}{dx}+y^{2}+x^{2}\frac{dy}{dx}+2xy=0$

Solve for dy/dx:

$\frac{dy}{dx}=\frac{-y(2x+y)}{x(x+2y)}$

Now, find the slope at x=3
• Nov 4th 2007, 08:04 PM
simply07
most excellent.. took me a while to understand what was going on but i got it now... however, i don't get how to find the slope!!! I can't isolate the y...
and i have to find where x=3...

if i try to plug it into the original equation - can't isolate y
when i try the derivative - still can't isolate y

how do i go about isolating the y because i get what to do afterwards it's this y that is not making sence to me!!! :confused:

thanks
• Nov 4th 2007, 08:25 PM
simply07
Nevermind
my friend cam by and i've seen the error in my ways...

thanks