$y = xy^2 + 4y^3 + 3x = 0$ at the point (1,-1)
To start I did the sum rule, separated all the terms, and got their derivatives
Then I expanded and substituted the point in and then used that slope to find the equation of the tangent line in point slope form
First, I was wondering if I did this correctly? Second, what is the slope of the tangent line?
Since there are y's on both sides of the equation, you need to use implicit differentiation, which essentially just boils down to treating y as a second variable and realizing that the derivative of y is $\frac{dy}{dx}$ not $\frac{dx}{dx} = 1$ like the derivative of x $$\frac {dy}{dx} = (y^2+2xy\frac {dy}{dx}) + 12y^2\frac {dy}{dx}+3$$ $$\frac {dy}{dx} - 2xy\frac {dy}{dx} - 12y^2\frac {dy}{dx} = y^2+3$$ $$\frac {dy}{dx}(1-2xy-12y^2)=y^2+3$$ $$\frac {dy}{dx} = -\frac {y^2+3}{2xy+12y^2-1}$$ The derivative at (1,-1) is then $$-\frac{(-1)^2+3}{2(1)(-1)+12(-1)^2-1}=-\frac 4 {9}$$ The slope normal line is the negative reciprocal of the derivative so it is therefore $\frac {9}{4}$