Question

Consider the curve given by the equation ln(y + x + 3) = xy + y3. Find an equation of the tangent line at the point (−4, 2)

Answer 1

ln(x+y+3) = xy + y^3
(1+y')/(y+x+3) = y + xy' + 3y^2 y'
Collect terms and solve for y':

y' =

     1-y(x+y+3)
--------------------------
(x+y+3)(x+3y^2)-1

so, at (-4,2), y' = -1/7

So, now you have a point and a slope, and the line is

y-2 = -1/7 (x+4)

Answer 2

I will assume that y3 is y^3

1/(y+x+3) * (dy/dx + 1) = x dy/dx + y + 3y^2 dy/dx
plug in the point (-4,2) :
1/(2-4+3) * (dy/dx + 1) = -4dy/dx + 2 + 12 dy/dx
1/ln1 * (dy/dx + 1) = 2 + 8dy/dx
arggghhh , 1/ln1 = 1/0, which is undefined

so we have a vertical tangent at (-4,2)
equation is x = -4

Answer 3

messed up in 4th step.

Answer 4

why is there a y' on the 3y^2 ?

Answer 5

@ steve

Answer 6

because y is a function of x. You must use the chain rule.

d/dx (3y^2) = d/dy (3y^2) * dy/dx