anonymous 5 years ago find the slope of the tangent line to the curve -1x^2-4xy-2y^3=92 at point (-2,-4)

1. anonymous

Use some implicit differentiation, differentiate the function, solve for dy/dx and plug in the points (-2,-4). Then, just use y-y1=m(x-x1) to find the tangent line.

2. anonymous

yes i get the concept and i did all that but, it's saying i have the wrong answer :(

3. anonymous

Hm...can you use the equation feature to clean up the function for me? Then I'll try it out and tell you what I get. :)

4. anonymous

i got -2x-4x y'+4x' y-6y^2 y'=0

5. anonymous

Mainly the first term...is it -x^2?

6. anonymous

yes but they wrote it as -1x^2

7. anonymous

i got y'=2x-4y/-4x-6y^2

8. anonymous

sot sure if that's right

9. anonymous

I'm getting this: $-2x - 4x*\frac{dy}{dx} - 4y - 6y^2\frac{dy}{dx} = 0$$-2x-4y = 4x \frac{dy}{dx} + 6y^2\frac{dy}{dx}$$\frac{dy}{dx} = \frac{-2x-4y}{4x+6y^2}.$

10. anonymous

sorry i couldn't read what u got

11. anonymous

...hm?

12. anonymous

sorry what u sent came up as a math processing error on my screen, but i g2g so thanks anyways

13. anonymous

maybe next time