[SOLVED] tangent line to the graph

**simply07** · November 4th 2007, 12:15 PM

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

**galactus** · November 4th 2007, 12:54 PM

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

**simply07** · November 4th 2007, 07:04 PM

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!!!

thanks

**simply07** · November 4th 2007, 07:25 PM

my friend cam by and i've seen the error in my ways...

thanks

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[SOLVED] tangent line to the graph

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