Question

Given 3xy-3(x+y)^3=-15 find the equation of the tangent line at the point (2,-3)

we have to first use implicit different but...:S

Answer 1

this is what i get:
\[\frac{dy}{dx} = \frac{y - 3(x+y)^{2}}{3(x+y)^{2} - x}\]
at point (2,-3)
dy/dx = slope = -6
y + 3 = -6(x -2)
y = -6x +9

Answer 2

using implicit differentiation...

Answer 3

yes, then use little algebra to isolate dy/dx

Answer 4

what'd you get for the implicit differentiation?

Answer 5

\[3(\frac{dy}{dx}x + y) - 9(x+y)^{2}(\frac{dy}{dx} +1) = 0\]

Answer 6

so how'd you get what you got up there?

Answer 7

move 2nd term over, divide by 3 on both sides
\[\frac{dy}{dx}x + y = 3(x+y)^{2}(\frac{dy}{dx} +1)\]
distribute right side
\[\frac{dy}{dx}x + y =3\frac{dy}{dx}(x+y)^{2} + 3(x+y)^{2}\]
get dy/dx 's on same side
\[y - 3(x+y)^{2} = 3\frac{dy}{dx}(x+y)^{2} - \frac{dy}{dx}x\]
factor out dy/dx, then divide

Answer 8

okay

Answer 9

i got it lol! thanks!1

Answer 10

:)