[Math] Implicit differentiation for $y\cos x = 4x^2 + 3y^2$

Question

I am stuck on doing this implicit differentiation problem below.
$$y \cos x = 4x^2 + 3y^2$$

I am now stuck at the following equality and I don't know how to proceed. Can someone help me?
$$y(−\sin x) + (\cos x)y' = 8x + 6yy' $$

Answer 1

What Marvis noted can be formulated below. Let $F(x,y)=0$ is a relation linking two variables $x$, $y$ implicitly. Then you can find $y'$ from this relation by calculating $$y'=\frac{-F_x}{F_y}$$ where in $F_x$ is differentiate of $F$ with respect to $x$. Now, $y \cos x = 4x^2 + 3y^2$ becames $y \cos x - 4x^2 - 3y^2=0$. Take $F(x,y)=y \cos x - 4x^2 - 3y^2=0$ and the rest is as Marvis did completely.