# Finding the tangent line and normal line to a curve

$y = xy^2 + 4y^3 + 3x = 0$ 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?

• You have to use implicit differentiation. – Natecat Dec 10 '16 at 22:08
• Why isn't the answer accepted? – Viktor Glombik Feb 8 at 15:31

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 $-\frac {9}{4}$