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)
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)
6 answers
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
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
messed up in 4th step.
why is there a y' on the 3y^2 ?
@ steve
because y is a function of x. You must use the chain rule.
d/dx (3y^2) = d/dy (3y^2) * dy/dx
d/dx (3y^2) = d/dy (3y^2) * dy/dx