1. ## isolating a variable...difficulties

Once again, you guys are helping me out heaps. Thanks! My maths teacher I believe has put many tricks into these questinos. I eventually need to find the gradient of this curve. I would normally isolate y, and then take the derivative. But I'm having issues isolating y. Any help is appreciated...thanks

$xy(x+y)+16=0$

2. Originally Posted by solarscott
Once again, you guys are helping me out heaps. Thanks! My maths teacher I believe has put many tricks into these questinos. I eventually need to find the gradient of this curve. I would normally isolate y, and then take the derivative. But I'm having issues isolating y. Any help is appreciated...thanks

$xy(x+y)+16=0$
that is the main reason for using implicit differentiation.

but, if you insist on isolating y ...

$xy^2 + x^2 y + 16 = 0$

$y = \frac{-x^2 \pm \sqrt{x^4 - 64x}}{2x}$

3. ahhh, I knew there was another way to do it...just forgot the name..implicit differentiation! Thanks. I will look that up...Im believe that is the way he is expecting it to be done...

4. It looks more like a FOIL problem to me...

xy(x+y)+16=0

x^2+y^2+2y+2x+16=0

5. Originally Posted by Caturdayz
It looks more like a FOIL problem to me...

xy(x+y)+16=0

x^2+y^2+2y+2x+16=0
it's not a FOIL problem.

6. Yep, I just worked it through to find what I was supposed to (the equation of the tangent where the gradient = -1, and the answer I think makes sense...

I found:
$y'=(-y^2-2xy)/(x^2+2xy)$

subbing in y'=-1, I solve for:

$-1=(-y^2-2xy)/(x^2+2xy)$
$-x^2-2xy=-2xy-y^2$
$x^2=y^2$
$y=-+x$

So the equation of the tangent where the gradient =-1 would be y=+-x? Does this make sense? Thanks